Continuous R-valuations

We introduce continuous $R$-valuations on directed-complete posets (dcpos, for short), as a generalization of continuous valuations in domain theory, by extending values of continuous valuations from reals to so-called Abelian d-rags $R$. Like the valuation monad $\mathbf{V}$ introduced by Jones and Plotkin, we show that the construction of continuous $R$-valuations extends to a strong monad $\mathbf{V}^R$ on the category of dcpos and Scott-continuous maps. Additionally, and as in recent work by the two authors and C. Th\'eron, and by the second author, B. Lindenhovius, M. Mislove and V. Zamdzhiev, we show that we can extract a commutative monad $\mathbf{V}^R_m$ out of it, whose elements we call minimal $R$-valuations. We also show that continuous $R$-valuations have close connections to measures when $R$ is taken to be $\mathbf{I}\mathbb{R}^\star_+$, the interval domain of the extended nonnegative reals: (1) On every coherent topological space, every non-zero, bounded $\tau$-smooth measure $\mu$ (defined on the Borel $\sigma$-algebra), canonically determines a continuous $\mathbf{I}\mathbb{R}^\star_+$-valuation; and (2) such a continuous $\mathbf{I}\mathbb{R}^\star_+$-valuation is the most precise (in a certain sense) continuous $\mathbf{I}\mathbb{R}^\star_+$-valuation that approximates $\mu$, when the support of $\mu$ is a compact Hausdorff subspace of a second-countable stably compact topological space. This in particular applies to Lebesgue measure on the unit interval. As a result, the Lebesgue measure can be identified as a continuous $\mathbf{I}\mathbb{R}^\star_+$-valuation. Additionally, we show that the latter is minimal.


Introduction
The probability of an event is most often than not understood as a real number between 0 and 1, and measures, as well as continuous valuations, take their values in R + , the set of non-negative real numbers extended with +∞.What is there that is so special with real numbers, and can we replace R + by some elements in some other structure?The question was once asked by Vincent Danos to the first author, and came back to the authors in an attempt to formulate an alternative to measures and continuous valuations with values taken as exact reals, in the sense of Real PCF [10,9,8,24] for example.Exact reals are modeled there as intervals that enclose the true value that is intended, and computation proceeds by refining these intervals further and further.Indeed, one of the points of this paper is that we can extend continuous valuations to an interval-valued form of continuous valuations, with an interval-valued integration theory.In addition, this also leads us to commutative valuations monads with intervals as values on the category of dcpos and Scott-continuous maps.
We should warn the reader that such an endeavor is probably useless for computation purposes.In the setting of type 2 theory of effectivity, Weihrauch has shown that, under reasonable assumptions, the map 2-3 σ-algebra is also generated by the intervals ]a, +∞] along (the Scott-open subsets, see below).Hence a measurable map h : X → R + is a map such that h −1 (]t, +∞]) ∈ Σ for every t ∈ R. Its Lebesgue integral can be defined elegantly through Choquet's formula: X hdµ def = +∞ 0 µ(h −1 (]t, +∞]))dt, where the right-hand integral is an ordinary Riemann integral.
This formula makes the following change-of-variables formula an easy observation: for every measurable map f : X → Y , for every measurable map h : The monotone convergence theorem states that, given any measure µ on a measurable space X, given any sequence (h n ) n∈N of measurable maps from X to R + that is pointwise monotonic, their pointwise supremum h is measurable, and X hdµ = sup n∈N X h n dµ.If (h n ) n∈N is antitonic instead, then a similar theorem holds provided that X h n dµ < +∞ for some n ∈ N (but not in general): the pointwise infimum h is measurable, and X hdµ = inf n∈N X h n dµ.
A practical way of building measures is Carathéodory's measure existence theorem, which is the following.A semi-ring R on X is a collection of subsets of X that is closed under finite intersections, and such that the complement of every element of R can be written as a finite disjoint union of elements of R. A map µ : R → R + is called σ-additive, extending the definition given above, if and only if for every countable (possibly empty) collection of pairwise distinct elements E n of R whose union E is also in R, µ(E) = n µ(E n ).Then µ extends to a measure on some σ-algebra containing R. A first use of this theorem is to establish the existence of Lebesgue measure λ on R, defined so that A measure µ on X is bounded if and only if µ(X) < +∞.A measure µ is σ-finite if there is a sequence measurable subsets of X whose union is X and such that µ(E n ) < +∞ for every n ∈ N. A π-system Π on a set X is a family of sets closed under finite intersections.If X is a measurable space such that Σ X = Σ(Π), any two σ-finite measures that agree on Π also agree on Σ X .In particular, Lebesgue measure on R is uniquely defined by the specification λ(]a, b[) = b − a.

Domain theory and topology.
A dcpo is a poset in which every directed family D has a supremum sup D. A prime example is IR ⋆ , the poset of closed intervals [a, b] with a, b ∈ R ∪ {−∞, +∞} and a ≤ b, ordered by reverse inclusion.Every directed family ([a i , b i ]) i∈I has a supremum i∈I [a i , b i ] = [sup i∈I a i , inf i∈I b i ].Among them, we find the total numbers a ∈ R ∪ {−∞, +∞}, which are equated with the maximal elements [a, a] in IR ⋆ .
Another example is R + , with the usual ordering.We will also consider IR ⋆ + , the subdcpo of IR ⋆ consisting of its elements of the form [a, b] with a ≥ 0.
We will also write ≤ for the ordering on any poset.In the example of IR ⋆ , ≤ is ⊇.The upward closure ↑ A of a subset A of a poset X is {y ∈ X | ∃x ∈ A, x ≤ y}.The downward closure ↓ A is defined similarly.A set A is upwards closed if and only if A = ↑ A, and downwards closed if and only if A = ↓ A. A subset U of a dcpo X is Scott-open if and only if it is upwards closed and, for every directed family D such that sup D ∈ U , some element of D is in U already.The Scott-open subsets of a dcpo X form its Scott topology.
The way-below relation ≪ on a poset X is defined by x ≪ y if and only if, for every directed family D with a supremum z, if y ≤ z, then x is less than or equal to some element of D already.We write ↑ ↑ x for {y ∈ X | x ≪ y}, and ↓ ↓ y for {x ∈ X | x ≪ y}.A poset X is continuous if and only if ↓ ↓ x is directed and has x as supremum for every x ∈ X.A basis B of a poset X is a subset of X such that ↓ ↓ x ∩ B is directed and has x as supremum for every x ∈ X.A poset X is continuous if and only if it has a basis (namely, X itself).A poset is ω-continuous if and only if it has a countable basis.Examples include R + , with any countable dense subset (with respect to its standard topology), such as the rational numbers in R + , or the dyadic numbers k/2 n (k, n ∈ N); or IR ⋆ and IR ⋆ + , with the basis of intervals [a, b] where a and b are both dyadic or rational.
We write OX for the lattice of open subsets of a topological space X.This applies to dcpos X as well, which will always be considered with their Scott topology.The continuous maps f : X → Y between two dcpos coincide with the Scott-continuous maps, namely the monotonic (order-preserving) maps that preserve all directed suprema.We write LX for the space of continuous maps from a topological space X to R + , the latter with its Scott topology, as usual.Such maps are usually called lower semicontinuous, or lsc, in the mathematical literature.Note that LX, with the pointwise ordering, is a dcpo.

2-4 Continuous R-valuations
There are several ways in which one can model probabilistic choice.The most classical one is through measures.A popular alternative used in domain theory is given by continuous valuations [18,17].A continuous valuation is a Scott-continuous map ν : OX → R + such that ν(∅) = 0 (strictness) and, for all U, V ∈ OX, ν(U ∪V )+ν(U ∩V ) = ν(U )+ν(V ) (modularity).There is a notion of integral x∈X h(x)dν, or briefly hdν, for every h ∈ LX, which can again be defined by a Choquet formula.The map h ∈ LX → hdν is Scott-continuous and linear.By definition, a linear map G : Conversely, any Scott-continuous linear map G : LX → R + is of the form h → hdν for a unique continuous valuation ν, given by ν(U 3 Rags, d-rags and continuous d-rags Definition 3.1 A rag is a tuple (R, 0, +, 1, ×) (or simply R) where (R, 0, +) is an Abelian monoid, (R, 1, ×) is a monoid, and × distributes over +.An Abelian rag is a rag whose multiplication × is commutative.
A semi-ring, or rig, is a rag which satisfies the extra law 0 × r = r × 0 = 0. R + , for example, is an Abelian rig, where + and × are as usual, modulo the convention that 0 × (+∞) = 0. We will see that IR ⋆ + is a rag, but not a rig.We also need some topological structure.
Definition 3.2 A d-rag is a rag R together with an ordering that makes it a dcpo, in such a way that + and × are Scott-continuous.
R + is a d-rag.In order to turn IR ⋆ + into a d-rag, we define its 0 element as [0, 0]; addition by The operations • ℓ and • r are product operations (for the ℓeft and right part, respectively), and are defined so that x • ℓ y and x • r y are equal to the usual product xy unless one of x, y is equal to 0 and the other is equal to +∞.We need two distinct, left and right, product operations in order to ensure Scott-continuity, as we now explain.We must define 0 • ℓ (+∞) (= (+∞) • ℓ 0) as 0, since 0 • ℓ (+∞) must be equal to sup r∈R + 0 • ℓ r = 0. Symmetrically, we must define 0 • r (+∞) (= (+∞) • r 0) as +∞, because 0 • r (+∞) must be equal to inf r>0 r • r (+∞) = +∞.With those choices, we have the following easily checked fact.

Continuous R-Valuations
Let R be a fixed Abelian d-rag.One might be tempted to define continuous R-valuations on a space X as Scott-continuous maps from OX to R satisfying some appropriate forms of strictness and modularity, but, as we have argued in the introduction, this is fraught with difficulties when the additive unit is not the least element of R.
Since continuous valuations on X correspond bijectively to linear Scott-continuous maps from LX to R + , another route is to define continuous R-valuations as certain maps from a variant of LX to R instead of R + .As we will see, this leads to a streamlined theory.
Given any space X, let L R X be the dcpo of all continuous maps from X to R, with the pointwise ordering.With pointwise addition and multiplication, L R X is also an Abelian d-rag.
We write V R X for the dcpo of all continuous R-valuations on X, with the pointwise ordering.Remark 4.2 When R = R + , L R X = LX, so that V R X can be equated with the dcpo VX of ordinary continuous valuations.

2-5
In order to help understand the definition, it is profitable to use the integral notation hdν to mean ν(h).Hence Definition 4.1 requires that (a × h)dν = a × hdν and (h + h ′ )dν = hdν + h ′ dν.
Beware that the constant 0 map from L R X to R is not a continuous R-valuation, unless R is a rig: homogeneity would imply 0 = a × 0, which fails in IR ⋆ + , for example.Also, we do not require ν(0) = 0 in Definition 4.1, where 0 is the constant 0 map.This would be a consequence of homogeneity if R were a rig.
Addition and multiplication by scalars in R are defined pointwise on V R X.This allows us to make sense of the following definition.
Our definition of the R-Dirac mass reads hdδ x = h(x) in integral notation.
Remark 4.4 A simple valuation on X is one of the form n i=1 r i δ x i , where n ∈ N, each point x i is in X, and each coefficient r i is in R + .While continuous valuations can be equated with continuous R-valuations with R = R + (Remark 4.2), simple valuations and elementary R-valuations are closely related but different concepts, even when R = R + .First, rδ x is an elementary R + -valuation even when r = +∞, but is a simple valuation only if r < +∞.Second, we require n ≥ 1 in the definition of elementary R-valuations, but n can be equal to 0 in the definition of a simple valuation.The reason why we require n ≥ 1 is that the constant 0 map is not a continuous R-valuation in general, as noticed above.
The inductive closure of a subset A of a dcpo Z is the smallest subset of Z that contains A and is closed under directed suprema.It is obtained by taking all directed suprema of elements of A, all directed suprema of elements obtained in this fashion, and proceeding this way transfinitely.
A pointed dcpo is one with a least element ⊥.
Proposition 4.5 Let R be an Abelian d-rag with a least element ⊥ that is absorbing for multiplication, viz., ⊥ × a = ⊥ for every a ∈ R. For every non-empty space X, the constant map ⊥ : h ∈ L R X → ⊥ is the least element of V R X, and also of V R m X.Thus, V R X and V R m X are pointed dcpos.
The last claim is obvious.✷ Proposition 4.5 applies to the case R = R + , where the bottom element is 0, and 0 × r = r for every r (including +∞).It also applies to the case R = IR ⋆ + , where the bottom element is [0, +∞], and again Remark 4.6 There is a very similar notion of integration of interval-valued functions, yielding interval values, due to Edalat [7], which he uses to define interval-valued integrals of measurable functions.The purpose is to set up a computable framework for Lebesgue measure and integration theory.The two integrals considered in [7] and in the present paper are similar, but different in a subtle way.There are small differences, such as the fact that Edalat allows one to integrate functions with values in IR, whereas we only integrate with values in IR ⋆ + , but the main difference is best illustrated by the following example.Let λ be Lebesgue measure on [0, 1], and h n : [0, 1] → IR ⋆ + map every x ∈ [0, 1/2 n ] to [0, ∞] and every x ∈]1/2 n , 1] to [0, 0].The maps h n form a chain whose supremum is the function h that maps every element of ]0, 1] to [0, 0] and 0 to [0, ∞].Using Edalat's integral, we have h n dλ = [0, ∞] for every n ∈ N, but hdλ = [0, 0].This shows that Edalat's integral is not a continuous IR ⋆ + -valuation in general.We will propose a way to fix this issue in Section 7.

Monads of continuous R-valuations
We fix an Abelian d-rag R. We will see that V R and V R m define strong monads on the category Dcpo of dcpos and Scott-continuous maps.This is essential in describing probabilistic effects, following Moggi's seminal work [22,23].We use Manes' presentation of monads [21]: a monad (T, η, † ) on a category C is a function T mapping objects of C to objects of C, a collection of morphisms η X : X → T X, one for each object X of C, and called the unit, and for every morphism f : X → T Y , a morphism f † : T X → T Y called the extension of f ; those are required to satisfy the axioms: Then T extends to an endofunctor, acting on morphisms through Proposition 5.1 The triple (V R , η, † ) is a monad on the category of dcpos and Scott-continuous maps, where η X : X → V R X maps x to δ x , and for every f : Proof.Verifying that η is Scott-continuous is routine.
Let us look at f † .For every ) is also Scott-continuous, since f (x) is Scottcontinuous for every x ∈ X and since ν is itself Scott-continuous.It is easy to see that f † (ν) is linear, too, because f (x) is linear for every x ∈ X, and because ν is linear.Hence f † (ν) is an element of V R Y for every ν ∈ V R X. Finally, f † itself is Scott-continuous, as one easily checks.
In integral notation, this means ν ′ def = V R (f )(ν) satisfies kdν ′ = (k • f )dν.This is a formula that is typical of the image measure of ν by f , where ν is a measure.We may think of V R (f )(ν) as the image of the continuous R-valuation ν by f .
We will now show that V R m defines a submonad of V R .To this end, we need to know more about inductive closures.A d-closed subset of a dcpo Z is a subset C such that the supremum of every directed family of elements of C, taken in Z, is in C. The d-closed subsets form the closed subsets of a topology called the d-topology [19, Section 5], and the inductive closure of a subset A coincides with its d-closure cl d (A), namely its closure in the d-topology.
We note that every Scott-continuous map is continuous with respect to the underlying d-topologies.This is easily checked, or see [19,Lemma 5.3].In particular: Lemma 5.4 For every space X, V R m X is closed under addition and multiplication by elements of R, as computed in the larger space V R X.
Proof.Let us deal with addition.Multiplication is similar.
For every elementary R-valuation µ, the map f µ : ν ∈ V R X → µ + ν is Scott-continuous, and maps elementary R-valuations to elementary R-valuations.By Fact 5.3 with It follows that for every minimal R-valuation ν, the map g : µ ∈ V R X → µ+ν = f µ (ν) maps elementary R-valuations to minimal R-valuations.We observe that g is also Scott-continuous.By Fact 5.3 with the same A as above, g maps all elements of cl d Proof.The only challenge is to show that, for every ν for every dcpo X, and every x ∈ X, whence the following.
Proposition 5. 6 The triple (V R m , η, † ) is a monad on the category of dcpos and Scott-continuous maps.
A tensorial strength for a monad (T, η, † ) is a collection t of morphisms t X,Y : X × T Y → T (X × Y ), natural in X and Y , satisfying certain coherence conditions (which we omit, see [23].)We then say that (T, η, † , t) is a strong monad.We will satisfy ourselves with the following result.By [23,Proposition 3.4], in a category with finite products and enough points, if one can find morphisms t X,Y for all objects X and Y such that t X,Y • x, ν = T ( x•!, id Y ) • ν, where !: Y → 1 is the unique morphism from Y to the terminal object, then the collection of those morphisms is the unique tensorial strength.The category of dcpos has finite products, and has enough points, and specializing this to the V R monad, we obtain the following., y)) define the unique tensorial strength for the monad (V R , η, † ).
Proof.The previous observation shows that we must define t X,Y by t X,Y (x, ν) It is enough to check that t X,Y is Scott-continuous.This follows from the fact that application (of Given a tensorial strength t, there is a dual tensorial strength t ′ , where The monad T is commutative when they coincide. Lemma 5.9 Two morphisms f, g : X → Y in Dcpo that coincide on A ⊆ X also coincide on cl d (A).
Those two quantities are equal when µ is an elementary R-valuation m i=1 r i × δ x i (m ≥ 1), since the first one is equal to λk. m i=1 r i × ν(λy.k(xi , y)), the second one is equal to λk.ν(λy.m i=1 r i × k(x i , y)), and since ν is linear. For which we recognize as the integral permutation property, obtained in the classical measure-theoretic case as a consequence of Fubini's theorem.Fubini's theorem is more general, and states the existence of a product measure.A similar fact follows from the above results, as noticed by Kock [20].We write ⊗ for the morphism In integral notation, we obtain the following form of Fubini's theorem: As an additional benefit, we obtain (for free!) that the map ⊗ : (µ, ν) → µ ⊗ ν is Scott-continuous.Remark 5.12 A similar Fubini-like theorem was already obtained by Jones [17] for arbitrary (subprobability) continuous valuations, but in the setting of continuous dcpos only.Whether the Fubini-like formula above holds for every pair of continuous valuations µ and ν on arbitrary dcpos is unknown.We note that the problem would be easily solved if all continuous valuations were minimal, but that is not the case, as is shown in the paper [13].

Continuous R-valuations and measures I: A brief viewpoint
We look at the special cases of continuous R-valuations when R is R + or IR ⋆ + , and we investigate their relations to measures.
When R = R + , this is simple: as noticed in Remark 4.2, we can equate continuous R-valuations with continuous valuations.Next, continuous valuations and measures are pretty much the same thing on ω-continuous dcpos, namely on continuous dcpos with a countable basis.This holds more generally on de Brecht's quasi-Polish spaces [5], a class of spaces that contains not only the ω-continuous dcpos from domain theory but also the Polish spaces from topological measure theory.One can see this as follows.In one direction, every measure µ on a hereditarily Lindelöf space X is τ -smooth [2, Theorem 3.1], meaning that its restriction to the lattice of open subsets of X is a continuous valuation.A hereditarily Lindelöf space is a space whose subspaces are all Lindelöf, or equivalently a space in which every family of open sets contains a countable subfamily with the same union.Every second-countable space is hereditarily Lindelöf, and that includes all quasi-Polish spaces.In the other direction, every continuous valuation on an LCS-complete space extends to a Borel measure [6,Theorem 1.1].An LCS-complete space is a space that is homeomorphic to a G δ subset of a locally compact sober space.Every quasi-Polish space is LCS-complete; in fact, the quasi-Polish spaces are exactly the second-countable LCS-complete spaces [6, Theorem 9.5].Remark 6.1 A continuous valuation µ on an LCS-complete space X may extend to more than one Borel measure.However, the extension is unique if µ is σ-finite, namely if there is a monotone sequence open subsets of X whose union is the whole of X, and such that µ(U n ) < +∞ for every n ∈ N. Indeed, any extension of µ will be σ-finite in the usual sense.We conclude since any two σ-finite measures that agree on all open sets (which form a π-system) must agree on the Borel σ-algebra.
We now look in detail at the more complex case R = IR ⋆ + .We use the following notation.Given any element x of IR ⋆ + or of IR ⋆ , we write x − and x + for its endpoints, viz., (We will write r.1 for the constant function with value r, in order to distinguish it from the scalar value r.)Given any IR ⋆ + -continuous valuation F on a space X, we also define F − (h) as F (h) − and F + (h) as Lemma 6.2 (The view from the left I) Let X be any topological space.For every continuous IR ⋆ +valuation F on X, for every h ∈ L IR ⋆ + X, F − (h) only depends on h − , not on h + .Moreover, there is a unique continuous valuation ν F on X such that, for every h ∈ L IR ⋆ + X, Proof.For the first part, it suffices to show that ).We note that the bottom element [0, +∞] of IR ⋆ + is multiplicatively absorbing: for every Next, [0, +∞] satisfies the following partial absorption law for addition: ), and the right-hand side does not depend on h + .
In order to show the second part of the lemma, it suffices to observe that the map f → F − ([f, +∞.1]) is linear and Scott-continuous, and is therefore the integral functional of a unique continuous valuation

2-10 Continuous R-valuations
There are many ways in which we can reconstruct a continuous IR ⋆ + -valuation from a continuous valuation, and here is the simplest of all.Lemma 6.3 (The view from the left II) Let X be any topological space.For every continuous valuation ν on X, there is a smallest continuous For the continuous IR ⋆ + -valuation just given, the view from the right, namely F + (h), is the constant +∞, for every integrand h, including for the constant zero map.This cannot be the integral of h with respect to any measure, since the integral of the zero map is always zero, with respect to any continuous valuation or measure.
One possible view of continuous IR ⋆ + -valuations F is that of the specification of some unknown measure.F − gives a continuous valuation that is a lower bound on that measure, while F + measures how precise that specification is.In this setting, the continuous IR ⋆ + -valuation F built in Lemma 6.3 is the least precise specification for ν.
On more special topological spaces, we will see that every measure has a much more precise specification, and that it is minimal.
7 Continuous R-valuations and measures II: Measures as continuous IR ⋆ + -valuations We will see that every non-zero, bounded τ -smooth measure µ on a coherent topological space X gives rise to a continuous IR ⋆ + -valuation in a natural way.(A measure µ on X is bounded if µ(X) < ∞, and we recall that it is τ -smooth if and only if it restricts to a continuous valuation on OX.)As a first step, we need to define integrals of functions with values in R + , not just R + , as is done classically.
More precisely, given a measure µ on a topological space X (with its Borel σ-algebra), we can define the Lebesgue integral x∈X f (x)dµ of any measurable map f : X → R + .We extend this definition to measurable maps f from X to R + .Just as with multiplication in rags, this comes in two flavors.
Perhaps the most natural extension is: It is known that the Lebesgue integral, as used on the right of (1) can be defined through the following, so-called Choquet formula [4, Chapter VII, Section 48.1, p. 265]: where the integral on the right is now an indefinite Riemann integral.As a consequence, and since (min(f ( ), r) −1 (]t, ∞]) is empty for every t ≥ r, and equal to f −1 (]t, ∞]) for every t < r, we can rewrite (1) as: We observe that this lower integral is linear and ω-continuous (by the monotone convergence theorem).It also commutes with the product structure of the d-rag R + .We also note the following change of variable formula, for future reference: Goubault-Larrecq, and Jia

2-11
for every measurable map j : X → Y , for every measurable map f : Y → R + , for every measure µ on X, and where j[µ] is the image measure, defined by j[µ](E) def = µ(j −1 (E)).This is an obvious consequence of (3).
A function f : X → R + is lower semicontinuous if and only if it is continuous from X to R + with the Scott topology; equivalently, for every r Every lower semicontinuous function is measurable.
2. For every dµ is equal to 0 by definition, and The new, key case is when a = ∞.We split this into two subcases.If µ(f −1 (]0, ∞])) = 0, namely if f is µ-a.e.zero, then min(f ( ), r) and min(∞ • ℓ f ( ), r) are also µ-a.e.zero, so We use the monotone convergence theorem, and the fact that sup ↑ n∈N min(f n (x), r) = min( sup ↑ n∈N f n (x), r) for all x ∈ X and r ∈ R + .4. Since Riemann integration of non-increasing maps from R + to R + is Scott-continuous (see for example Lemma 4.2 in [26]), it follows from (3) that the lower integral x∈X f (x)dµ is Scott-continuous in the lower semicontinuous map f , provided that µ is τ -smooth.✷ We also consider the following, upper integral.This will really only make sense when the integrated function f is upper semicontinuous, namely when for every r ∈ R + \ {0}, f −1 ([0, r[) is open in X; and when the measure µ is non-zero (µ(X) = 0), and τ -smooth.
A support of a measure µ on X is any set E such that, for all measurable subsets A and B of X such that A ∩ E = B ∩ E, µ(A) = µ(B).When E is itself measurable, this is equivalent to requiring µ(E) = µ(X), and when µ is additionally bounded (i.e., µ(X) < ∞), this is equivalent to µ(X \ E) = 0. We sometimes say that µ is supported on E to mean that E is a support of µ.

2-12
Continuous R-valuations For every τ -smooth measure µ on X, the intersection of all closed supports of µ is again a closed support of µ: this smallest closed support will be denote as supp µ.But beware that there might be smaller (non-closed) supports.For example, supp (δ x ) is equal to the closure ↓ x of the point x, but {x} is a smaller (non-closed) support.Note that {x} is the intersection of the compact saturated set ↑ x, which happens to be a support of µ = δ x , with supp (µ).
In general, not all compact saturated sets Q are measurable, so we will restrict to measurable compact saturated subsets in the sequel.The intersection of two supports E and E ′ may also fail to be a support, but if one of them (say E ′ ) is measurable, then E ∩ E ′ is also a support.(Indeed, let A, B be measurable . Since E ′ is a support of µ, and since A and A ∩ E ′ have the same intersection with E ′ , µ(A ∩ E ′ ) = µ(A), and similarly µ(B ∩ E ′ ) = µ(B).Therefore µ(A) = µ(B).) For every τ -smooth measure µ on X, we say that a measurable map f : X → R + is µ-bounded if and only if there is a measurable compact saturated support Q of µ such that f is bounded on Q ∩ supp µ, namely if sup x∈Q∩supp µ f (x) < ∞.We will also say that Q is a witness of µ-boundedness of f , or that f is µ-bounded, witnessed by Q, in that case.We say that f is µ-unbounded if it is not µ-bounded.Lemma 7.2 For every non-zero τ -smooth measure µ on a topological space X, for every compact saturated support Q of µ, Q ∩ supp µ is non-empty.
Proof.Otherwise, supp µ and the empty set have the same intersection with Q, and since Q is a support of µ, we would have µ(supp µ) = µ(∅) = 0. Since supp µ is a measurable support of µ, µ(supp µ) = µ(X), and therefore we would have µ(X) = 0, contradicting the fact that µ is non-zero.✷ Lemma 7.3 For every non-zero τ -smooth measure µ on a topological space X, and for every upper semicontinuous map f : Proof.Every upper semicontinuous R + -valued function f reaches its maximum on any non-empty compact set K.Here is a quick proof: let a def = sup x∈K f (x), and assume that f (x) < a for every x ∈ K.The open sets f −1 ([0, r[) with r ∈ [0, a[ form an open cover of K. We extract a finite subcover f −1 ([0, r[), where r ranges over some finite set A of numbers strictly below a.This implies that, for every x ∈ K, f (x) < r for some r ∈ A, so that a = sup x∈K f (x) < max A < a, a contradiction.
We now apply this to K def = Q ∩ supp µ, which is non-empty by Lemma 7.2.✷ Corollary 7.4 For every non-zero τ -smooth measure µ on a topological space X, for every upper semicontinuous map f : X → R + , f is µ-unbounded if and only if for every measurable compact saturated support This being done, for a τ -smooth measure µ and an upper semicontinuous map f : X → R + , we define: We say that a topological space is coherent if and only if the intersection of any two compact saturated subets is compact (and saturated).Lemma 7.5 Let µ be a non-zero τ -smooth measure on a topological space X.The upper integral (5) is: (ii) • r -homogeneous: for every upper semicontinuous map f : (iii) Scott-cocontinuous if µ is also bounded: for every filtered family (f i ) i∈I of upper semicontinuous maps from X to R + , 2-13 (iv) above the lower integral: for every measurable map g : X → R + , for every upper semicontinuous map f : Proof.We prove item 5 first.When g is µ-bounded, witnessed by Q, we can define a new measurable map g.1 Q∩supp µ , which maps every x ∈ Q ∩ supp µ to g(x), and all other points to 0. Then g.1 Q∩supp µ is bounded, and coincides with g on Q ∩ supp µ.Since the latter is a support of µ, it is an easy exercise, using (3), to show that − x∈X g(x)dµ is equal to − x∈X g(x).1 Q∩supp µ (x)dµ, which is the ordinary Lebesgue integral x∈X g(x).1 Q∩supp µ (x)dµ.
1.If f and g are both µ-bounded, witnessed respectively by Q and Q ′ , then so is f + g, witnessed by Q ∩ Q ′ .The latter is measurable, and compact saturated since X is coherent.It is also a support of µ, since Q (or Q ′ ) is measurable.The claim then follows from Lemma 7.1, item 1.
If, say, f is not µ-bounded, then for every measurable compact saturated support Q of µ, there is a point x ∈ Q ∩ supp µ such that f (x) = ∞ by Corollary 7.4.Then, f (x) + g(x) is also equal to ∞, showing that f + g is not µ-bounded either.In particular, + x∈X (f (x) + g(x))dµ and 2. If f is µ-bounded and a = ∞, then a • r f is also µ-bounded: for every measurable compact saturated support Q of µ, f and therefore a • r f is bounded on Q ∩ supp µ.Then the claim follows from Lemma 7.1, item 2, and the fact that • ℓ and • r both coincide with the ordinary product on R + .
If a = ∞, then by definition ∞• r The latter equality follows from the fact that the constant map ∞ is not µ-bounded; indeed, for every measurable compact saturated support Q of µ, Q ∩ supp µ is non-empty by Lemma 7.2, so that ∞ is not bounded on that set.
If f is not µ-bounded but a ∈ R + , then for every measurable compact saturated support Q of µ, there is a point x ∈ Q ∩ supp µ such that f (x) = ∞ by Corollary 7.4.Then a • r f (x) = ∞ as well.This shows that a • r f is not µ-bounded either.It follows that i∈I f i of upper semicontinuous maps f i 's is upper semicontinuous.Let us write i j if and only if f i ≤ f j .
If f i 0 is µ-bounded for some i 0 ∈ I, then f i ≤ f i 0 is also µ-bounded for every i i 0 , and witnessed by the same measurable compact saturated set Q. Similarly, f is also µ-bounded, witnessed by Q.We let r be an upper bound of f i 0 on Q ∩ supp µ.Then, using item 5, Indeed, the map (r − f ( ))1 Q∩supp µ also takes its values in R + , and the sum of x∈X f (x)1 Q∩supp µ (x)dµ and of x∈X (r − f (x))1 Q∩supp µ (x)dµ is equal to x∈X r1 Q∩supp µ (x)dµ = rµ(Q ∩ supp µ) = rµ(X), since Q ∩ supp µ is a measurable support of µ.Since integration of lower semicontinuous maps with respect to

2-14
Continuous R-valuations a τ -smooth measure is Scott-continuous, as in Lemma 7.1, item 4, we obtain: If no f i is µ-bounded, then f cannot be µ-bounded either, as we now claim.If f is µ-bounded, witnessed by Q, then, let r ∈ R + be such that for every The intersection of a compact set and of a closed set is compact, so Q ∩ supp µ is compact, and therefore (f i∈I is a directed family, we can assume that this subcover consists of just one open set f −1 i ([0, r[).But that implies that f i is bounded on Q ∩ supp µ, hence µ-bounded, a contradiction.
Hence we have proved that f is not µ-bounded, so We let R def = IR ⋆ + , and we fix a topological space X.For every h ∈ L R X, for every x ∈ X, h(x) is an interval [h − (x), h + (x)].The function h − is lower semicontinuous.Indeed, for every r ∈ R + \ {0}, if and only if r < a. Symmetrically, the function h + is upper semicontinuous.Our preparatory steps on the lower and upper integrals then allow us to make sense of the following definition.The fact that dµ is by Lemma 7.5, item 4. Definition 7.6 For every τ -smooth measure µ on a topological space X, we define µ : Proposition 7.7 For every non-zero, bounded τ -smooth measure µ on a coherent topological space X, µ is a continuous IR ⋆ + -valuation.
Proof.First, µ is linear by Lemma 7.1 (items 1 and 2) and Lemma 7.5 (items 1 and 2).We verify that It it Scott-continuous.Let (h i ) i∈I be a directed family in L IR ⋆ + X, with supremum h.We aim to show that µ(h) = sup ↑ i∈I µ(h i ).On the one hand, dµ by Lemma 7.1, item 4. On the other hand, h + = inf ↓ i∈I h + i , so dµ by Lemma 7.5, item 3. ✷ Remark 7.8 We think of µ as being really the measure µ, seen as a continuous IR ⋆ + -valuation.Note in particular that for every bounded continuous map h : Goubault-Larrecq, and Jia

2-15 8 Continuous R-valuations and measures III: continuous IR ⋆ + -valuations as approximations of measures
Let us say that [a, b] approximates x if and only if a ≤ x ≤ b, and that a continuous map h ∈ L IR ⋆ + X approximates a measurable map f : X → R + if and only if h(x) approximates f (x) for every x ∈ X.
We will say that a continuous IR ⋆ + -valuation ν approximates a measure µ on X if and only if, for every measurable map f : X → R + and for every h ∈ L IR ⋆ + X that approximates f , ν(h) approximates Lemma 8.1 For every non-zero, bounded τ -smooth measure µ on a coherent topological space X, µ approximates µ.
Proof.We consider any measurable map f : X → R + and any h ∈ L IR ⋆ + X that approximates f .We write h(x) as [h − (x), h + (x)] for every x ∈ X, so that where the last inequality is by Lemma 7.5, item 4 applied to f ≤ h + .✷ Our objective is now to show that µ is the most precise, namely the largest, continuous IR ⋆ + -valuation that approximates µ, under some reasonable assumptions.This will notably hold when the ambient space X is compact Hausdorff and second-countable, for example [0, 1] with its usual, metric topology.More generally, this will hold when X is stably compact, second-countable, and contains a sufficiently nice support K of µ.
The value of restricting to second-countable spaces is the following.
Lemma 8.2 Let X be a topological space, and B be a base of its topology that is closed under finite unions.
(i) For every compact saturated subset Q of X, and every open neighborhood U of Q, there is a V ∈ B such that Q ⊆ V ⊆ U .(ii) Every compact saturated subset of X is equal to the intersection of the sets in B that contain it.
In particular, if X is second-countable, then every compact saturated subset of X is measurable.
Proof.(i) We write U as the union of the sets V ∈ B that are included in V .This forms an open cover of Q, from which we can extract a finite subcover.Since B is closed under finite unions, there is a V ∈ B that contains Q and is included in U .
(ii) Let Q be compact saturated in X.Since Q is saturated, Q is the intersection of its open neighborhoods U .Then claim 2 follows from 1.
When B is countable, Q is then a countable intersection of open sets, so Q is measurable.✷ A space X is stably compact if and only if it is sober, locally compact, compact, and coherent.We let X patch denote X with its patch topology, which is the smallest topology that contains the original open subsets of X and the complements of compact saturated subsets of X.When X is stably compact, and ≤ is its specialization ordering, (X patch , ≤) is a compact pospace, meaning that X patch is compact Hausdorff, and that the graph of ≤ is closed in X patch × X patch .We say that a subset of X is patch-open if it is open in X patch .Similarly, we use the terms patch-closed, patch-compact.If X is stably compact, then patch-closed and patch-compact are synonymous.We should add that the original open subsets of X can be recovered as those patch-open subsets that are upwards-closed with respect to ≤. Reasoning similarly, the larger dcpo IR ⋆ of all closed intervals in R ∪ {−∞, ∞}, ordered by reverse inclusion, is also a bc-domain, hence is also stably compact.(To make it clear, note that

2-16
Continuous R-valuations Both bc-domains are second-countable as well.Indeed, as continuous dcpos, they have a basis B of intervals with rational endpoints, and then the set of Scott-open sets ↑ ↑ b, b ∈ B, forms a countable base of the Scott topology.Patch-compact subsets K of stably compact subspaces X enjoy many nice properties.For example, their downward closure ↓ K in X is closed [12,Exercise 9.1.43].In fact, we have the following, where K is order-convex if and only if for all x, y, z such that y ≤ x ≤ z, if y, z ∈ K then x ∈ K. Lemma 8.4 For every compact, order-convex subset K of a stably compact space X, K is patch-compact if and only if ↓ K is closed.In that case, ↓ K is the closure cl(K) of K in X, and K = ↑ K ∩ ↓ K.
Proof.We only have to show that if K is compact and order-convex and if ↓ K is closed, then it is patchcompact.This will be a consequence of the last equality K = ↑ K ∩ ↓ K, since ↑ K is compact saturated, and we have assumed that ↓ K is closed, hence both are patch-closed; then K is patch-closed, too, hence patch-compact.
The inclusion K ⊆ ↑ K ∩ ↓ K is clear.Conversely, every x ∈ ↑ K ∩ ↓ K is such that y ≤ x ≤ z for some y, z ∈ K, so order-convexity implies y = z, and therefore also x = y = z.In particular, x is in K. ✷ We will say that a subset K of a space X is Hausdorff if and only if it is Hausdorff as a subspace, namely with the subspace topology inherited from X. Since the specialization ordering of a Hausdorff space is equality, every Hausdorff subset is trivially order-convex.
When K is patch-compact in a stably compact space X, we have the serendipitous property that K, with the subspace topology, is stably compact, and that the patch topology on K is the subspace topology inherited from X patch [12, Proposition 9.3.4];also, the specialization ordering on K is the restriction of that on X.
The previous remark, together with the fact that K patch = K if K is compact and Hausdorff (since every compact set is already closed in K), entails the following.Lemma 8.6 Let X be a stably compact space, and K be a Hausdorff, patch-compact subset of X.Then K has both the subspace topology inherited from X, and the subspace topology inherited from X patch .✷ We also note that every Hausdorff subset is order-convex, since the specialization ordering on any Hausdorff space is equality.Proposition 8.7 Let K be a Hausdorff, patch-compact subset of a stably compact, second-countable space X.Let µ be a non-zero measure on X supported on K, and ν be a continuous IR ⋆ + -valuation on X.If ν approximates µ, then ν ≤ µ.
Proof.We first note that, since X is second-countable, every measure on X is τ -smooth, in particular µ.
Let h be an arbitrary continuous map in

2-17
For the second claim, we distinguish two cases.If h + is µ-bounded, then since h approximates h + , ν(h) approximates − x∈X h + (x)dµ, which is equal to + x∈X h + (x)dµ by Lemma 7.5, item 5.In particular, The only case where we have to work a bit is the final case, when h + is µ-unbounded.We let Q def = ↑ K.This is a compact saturated subset of X, and since X is second-countable, Q is measurable by Lemma 8.2.Moreover, Q is a support of µ, since Q contains K, which is already a support of µ.
Q ∩ supp µ is then a measurable, compact support of µ.We claim that supp µ is included in the closure cl(K) of K in X. Equivalently, we claim that every open set U that intersects supp µ also intersects K. Since U intersects supp µ, by definition of supp µ, we have µ(U ) > 0. If it did not intersect K, then U and the empty set would have the same intersection with K, and that would imply µ(U ) = µ(∅) = 0, which is impossible.
Since supp µ ⊆ cl(K), we obtain that Since K is compact Hausdorff hence locally compact, x 0 has a base of compact neighborhoods we can write it as C ∩ K for some closed subset C of X, and then C ⊇ C i , so The family (K i ) i∈I is filtered: for all i, j ∈ I, K i ∩ K j is a compact neighborhood of x 0 , using the fact that the intersection of two compact sets in a Hausdorff space is compact; hence This is a closed subset of X containing x 0 .We claim that C is exactly the downward closure of x 0 in X.It only remains to show that C ⊆ ↓ x 0 .Let us assume the contrary: for some is open in X patch , and therefore (X \ ↑ x) ∩ K is open in K, using Lemma 8.6.Since (K i ) i∈I is a base of neighborhoods of x 0 in K, some K i is included in (X \↑ x)∩ K.This is impossible, since x ∈ ↓ K i .
For every i ∈ I, let h i be the function that maps every x ∈ C i to ∞, and all other points to h + (x).This is an upper semicontinuous map, since h −1 i ([0, r[) = h + −1 ([0, r[) \ C i for every r ∈ R + .The family (h i ) i∈I is filtered, since (C i ) i∈I is a filtered family of sets.Moreover, for every x ∈ X, inf ↓ i∈I h i (x) = h + (x).If x ≤ x 0 , we argue as follows.First, so x is not in C i for some i ∈ I, and therefore h i (x) = h + (x).This implies that inf ↓ i∈I h i (x) ≤ h + (x), while the reverse inequality is obvious.
).However, we claim that the left-hand side is equal to ∞, so that ν We have an open set U i that intersects supp µ (at x 0 ): by definition of the support, µ(U i ) > 0. (Namely, if we had µ(U i ) = 0, then U i would be included in the largest open subset with zero µ-measure, which is the complement of supp µ by definition.)It follows that Hence we have shown that ν + ([h − , h i ]) = ∞ for every i ∈ I. Taking suprema, and recalling that ), by the following argument, whose details we leave to the reader.If λ were minimal, then its view from the left would be in V R + m ([0, 1]), so λ would be a minimal valuation.Any minimal valuation is point-continuous, in the sense of Heckmann [15], because every simple valuation is point-continuous, and point-continuous valuations are closed under directed suprema.However, a valuation ν is point-continuous if and only if for every open set U , for every real number r such that 0 ≤ r < ν(U ), there is a finite subset A of U such that ν(V ) for every open neighborhood V of A; and λ fails to have this property, since every finite subset has open neighborhoods of arbitrarily small λ-measure.
Instead, we consider the image measure j[λ] on IR ⋆ , where j is the usual embedding of [0, 1] inside IR ⋆ , mapping x to the interval [x, x].We will show that, contrarily to λ, j[λ] is minimal.
This may sound somewhat paradoxical, considering that both have the same effect: drawing an interval at random with respect to measure j[λ] means drawing an interval of the form [x, x] with x ∈ [0, 1] with probability 1, where x is drawn uniformly at random in [0, 1], hence works just like λ; only the ambient space differs (IR ⋆ instead of [0, 1]).
We will say that j[λ] is the Lebesgue valuation on the unit interval in IR ⋆ .By Theorem 8.8 with is the continuous IR ⋆ + -valuation that approximates j[λ] in the most precise possible way.It is a bounded, non-zero, and τ -smooth measure (because IR ⋆ is second-countable, see Example 8.3).The objective of this section is to show that j To this end, we will show the stronger statement that j[λ] is the directed supremum of a countable chain of simple n ] -which we will write more simply as The second line is justified by the fact that × distributes over +.The inequality on the third line follows from the fact that n+1 ], and that h, product and addition are monotonic.The last line follows by rearranging the sum.✷ Goubault-Larrecq, and Jia

2-19
The chain (λ n ) n∈N then has a supremum in ) by definition of the latter, since every λ n is simple.Definition 9.2 Let λ be sup ↑ n∈N λ n .
In particular, j Proof.The interval [0, 1], with its usual ordering, is a continuous dcpo, and its way-below relation ≪ is such that x ≪ y if and only if x = 0 or x < y.For every n ∈ N, for every x ∈ [0, 1], let j − n (x) be the largest integer multiple of 1  2 n way-below x; explicitly, j − n (x) both taken with their usual orderings).This implies that j n is lower semicontinuous from [0, 1] to R ∪ {−∞, ∞}, with their usual, Hausdorff topologies.

Let also j
. The function j + n is an upper semicontinuous map, and j − n (x) ≤ x ≤ j + n (x) for every x ∈ X.This implies that, if we define j n (x) as [j − n (x), j + n (x)], j n is a continuous map from [0, 1] to IR ⋆ , and j n ≤ j.
One checks easily that j − n ≤ j − n+1 , hence j + n ≥ j + n+1 , and therefore j n ≤ j n+1 .It follows that (j n ) n∈N is an increasing chain of continuous maps.Moreover sup ↑ n∈N j n = j.Indeed, for every x ∈ X, sup ↑ n∈N j − n (x) = x, and inf ↓ n∈N j Note that h − is lower semicontinuous, or equivalently Scott-continuous from IR ⋆ to IR ⋆ + .We have: h − (j(x))dλ by the change of variable formula (4) h − (j n (x))dλ by Lemma 7.1, item 3 (or 4).
The function h − • j n is piecewise constant: it takes the value h Since the values in [0, 1] that are not in one of those open subintervals, namely the integer multiples of 1  2 n , form a set of Lebesgue measure 0, they do not contribute to the integral, so: and we recognize the left endpoint of the final interval of ( 6).

2-20
Continuous R-valuations We now deal with the right endpoint.For that, we need to understand what the supports of j[λ] are on IR ⋆ .Let K be the image of [0, 1] by j in IR ⋆ .We have seen in Example 8.5 that ↓ K is closed in IR ⋆ , hence that K is patch-compact in IR ⋆ , using Lemma 8. 4.
For every open subset U of IR ⋆ , j[λ](U ) > 0 if and only if λ(j −1 (U )) > 0, if and only if j −1 (U ) is non-empty.Indeed, the Lebesgue measure of any non-empty open set is non-zero.Now j −1 (U ) is empty if and only if U does not intersect K, if and only if U does not intersect the closure of K, which is ↓ K, since ↓ K is closed.This entails that supp j[λ] = ↓ K.
We note that K is a support of j[λ].The easy argument is as follows.First, K is a set of maximal elements of IR ⋆ , so K = ↑ K; by Lemma 8.2, it is measurable.In order to show that K is a support of j[λ], it then suffices to observe that j[λ](K) = 1, and this follows from the fact that j[λ](K) = λ(j −1 (K)) = λ([0, 1]) = 1.
We claim that every compact saturated support Q of j[λ] must contain K.We argue as follows.By Lemma 8.6, and since Q is patch-closed in X patch (where this is a non-empty open subset of K, and therefore j −1 (U ) is a non-empty subset of [0, 1].Hence λ(j −1 (U )) > 0, so j[λ](U ) > 0. It follows that j[λ](Q ∩ K) = 1 − j[λ](U ) < 1.Since K is a support of j[λ], we obtain that j[λ](Q) = j[λ](Q ∩ K) < 1, and that contradicts that Q is a support of j[λ].
It follows that ↑ K (= K) is the smallest compact saturated support of j[λ].In that case, and with , the definition of µ-boundedness simplifies: h + is µ-bounded if and only if h + is bounded on It is even easier to show that h + • j is λ-bounded if and only if it is bounded on [0, 1].Indeed, supp λ = [0, 1], and therefore the only compact (hence closed) support Q of λ is [0, 1], so that there is only one possible set Q ∩ supp λ to be considered, namely [0, 1].
It remains to deal with the case where h + is µ-bounded, and we have seen that this means that h + is bounded on K. Let r ∈ R + be such that for every y ∈ K, h + (y) < r.Hence K is included in the open set h + −1 ([0, r[).
For every n ∈ N, let }}, a compact saturated set that contains K. (The upward closure of any finite set is compact saturated.)It is also easy to see that any point of ↓ n∈N Q n is of the form [x, x] with x ∈ [0, 1], so ↓ n∈N Q n = K.Since IR ⋆ is sober, it is well-filtered, and therefore K ⊆ h + −1 ([0, r[) implies the existence of an n ∈ N such that Q n ⊆ h + −1 ([0, r[).(Here is an alternate argument that avoids well-filteredness.Q n is compact saturated hence closed in IR ⋆patch .The complements of the sets Q n then form an open cover of the complement of h + −1 ([0, r[) in IR ⋆ .That complement is closed hence compact in IR ⋆patch .We can then extract a finite subcover, and since the sets Q n form a chain, there is a single n ∈ N such that the complement of Q n contains the complement of h + −1 ([0, r[).) Goubault-Larrecq, and Jia

2-21
Let n 0 be the natural number n that we have just found.Then we perform the following computation: h + (j(x))dλ.
The latter is justified by the fact that h + • j is bounded (by r) on [0, 1], hence is λ-bounded, as we have seen above.
Let us proceed.The first step below is justified by the fact that h + is upper semicontinuous, hence Scott-continuous from IR ⋆ to R + with the opposite ordering; in particular, h + maps directed suprema to filtered infima: h + (j n (x))dλ, where we have restricted the infimum to the indices above n 0 in the last line.(The infimum of a chain coincides with the infimum of any coinitial chain.)For every n > n 0 , for every x ∈ [0, 1], j n (x) is an interval of the form [ i−1 2 n , i 2 n ] (if x is in the interval ] i−1 2 n , i 2 n [, or if x = 0, or if x = 1), or of the form [ i−1 2 n , i+1 2 n ] (if x is exactly i 2 n , 1 ≤ i ≤ 2 n − 1).Since n > n 0 , whichever the case is, j n (x) is in Q n 0 , hence in h + −1 ([0, r[).This means that h + • j n is a bounded function, and therefore that + x∈[0,1] h + (j n (x))dλ is the ordinary Lebesgue integral x∈[0,1] h + (j n (x))dλ.Since h + • j n is piecewise constant (as with h − • j n , earlier on), that Lebesgue integral is equal to 2 n i=1 1 n ]).Therefore: and we recognize the right endpoint of the final interval of (6).✷

Conclusion
We have proposed an extension of the notion of continuous valuation, or measure, with values in suitable domains beyond R + .We have argued that continuous R-valuations, where R is a so-called Abelian d-rag, provide such an extension.Beyond R + , a particularly interesting Abelian d-rag is the domain of intervals IR ⋆ + , and we have shown that there is an ample supply of continuous IR ⋆ + -valuations stemming from measures.
There are many pending questions.For example, is V R (X) a continuous dcpo, provided that X is a continuous dcpo and R is a continuous Abelian d-rag?Is there a form of the Fubini theorem for continuous R-valuations, beyond the one we have obtained for minimal R-valuations?None of the usual proof arguments, in realms of measures or of continuous valuations, seems to apply.
where E is any measurable support of µ, − x∈X g(x)dµ ≤ + x∈X f (x)dµ; (v) for every µ-bounded upper semicontinuous map g : X → R + , witnessed by Q, − x∈X g(x)dµ = + x∈X g(x)dµ is the usual Lebesgue integral x∈X g(x)1 Q∩supp µ (x)dµ; dµ is larger than or equal to x∈X g(x).1 E∩Q∩supp µ (x)dµ, and the latter is equal to − x∈X g(x)dµ by a similar argument.If f is not µ-bounded, then + x∈X f (x)dµ = ∞, and the claim is trivial.✷

Example 8 .
3 IR ⋆ + is stably compact in its Scott topology.Indeed, it is a continuous dcpo in which any pair of elements [a, b] and [c, d] with an upper bound (namely, such that [a, b] ∩ [c, d] = ∅, or equivalently max(a, c) ≤ min(b, d)) has a least upper bound (which is [max(a, c), min(b, d)]).That kind of continuous dcpo is called a bc-domain, and every bc-domain is stably compact [12, Fact 9.1.6].