Fubini’s Theorem for Daniell Integrals

We show that in the theory of Daniell integration iterated integrals may be formed without exceptions. The order of integration may be interchanged, too. As a consequence, the Fubini theorem holds in full generality, thus covering the well-known Fubini-Stone theorem.


Introduction and main results
The Daniell integral is considered to be a particulary elegant approach to integration.Dating from 1918 [2], it took, nonetheless, several decades, until multiple Daniell integration had been treated in the literature.Two publications stand out.Stone [11] showed in 1948 for an integral on a product space, which can be reduced to an iterated integral, that its extension to summable functions may as well be expressed by an iterated integral.By contrast, Fremlin's far reaching results from 1972 [4] allow the construction of a product of integrals without touching iterated integrals.In this note we like to meet both these aspects and to proceed to a Fubini type theorem in such generality, as it is long-established in the narrower domain of Lebesgue integration.Our main observation is that iterated integrals may be formed for any two Daniell integrals, and that the order of integration may always be interchanged.
Let us recall some major notions from Daniell integration theory.Let F (Z) be the set of all function from some non-empty set Z into the real numbers.Consider a set L ⊂ F (Z), L = ∅, and a mapping I : are fulfilled, then one calls L a Riesz space (vector lattice) over Z and I an elementary Daniell integral on L.Then, together with f and g also sup(f, g) and inf(f, g) belong to L. (Here all operations or relations of functions are pointwisely defined.)The smallest Riesz space within F (Z) containing some set of functions S ⊂ F (Z) is denoted by L(S).We take the liberty to call an elementary Daniell integral I briefly an integral.We recall from the celebrated integration theory of Daniell, as presented e.g. in [1,6,10,12] that any integral I : L → R may be extended to the integral I : L 1 → R, where L 1 ⊃ L is the Riesz space of all real-valued I-summable functions.
To this extension we may apply the theorems of monotone and dominated convergence, as known from Lebesgue integration (see e.g.[10,12]).(As customary, we denote the integral and its extension by one and the same symbol.)Now, let us consider two integrals J : G → R and K : H → R with Riesz spaces G and H over X and Y , respectively.As indicated above, in order to combine them there are two aspects.First, by analogy to Lebesgue product measures, one asks for a product integral for all (x, y) ∈ X × Y .The natural candidate for L is the space L(G 1 ⊗ H 1 ), the smallest Riesz space over X × Y containing the tensor product G 1 ⊗ H 1 (the vector space spanned by all g ⊗ h with g ∈ G 1 and h ∈ H 1 ).This Riesz space L serves as the Riesz tensor product of G 1 and H 1 (characterized by an universality property, see [4]).
Second, one would like to express product integrals by iterated integrals.Let us consider for some function f ∈ F (X × Y ) and the derived functions given by K(f )(x) := K(f x ), J(f )(y) := J(f y ).
Then the iterated integrals J(K(f )) and K(J(f )) are well-defined.
Theorem 1.For any two integrals J : Moreover, any f ∈ L(G 1 ⊗ H 1 ) meets the conditions (a) and (b), and we have This integral I is written I = J ⊗ K and called the product integral of J and K.
With this theorem at hand, it is straightforward to proceed to a related Fubini type result.To this end, we somewhat weaken the requirements (a) and (b) and consider for f ∈ F (X × Y ) the conditions (a') there are a J-null set A ⊂ X and a K-null set B ⊂ Y such that f x ∈ H 1 and f y ∈ G 1 for x / ∈ A and y / ∈ B, (b') there are functions g ∈ G 1 and h ∈ H 1 such that g(x) = K(f x ) and h(y) = J(f y ) for x / ∈ A and y / ∈ B.
We recall that, for an integral I on the Riesz space L over Z, a set Note that the above functions g and h are in general not uniquely determined, yet the values of J(g) and K(h) are unique.Thus, it is justified to write K(f ) and J(f ) instead of g and h, respectively, and to build the iterated integrals J(K(f )) and K(J(f )) for functions f satisfying (a') and (b').Now our Fubini type theorem reads as follows.
Theorem 2. Let I : L → R be the product integral of J : G → R and K : H → R. Then any function f ∈ L 1 fulfils the conditions (a') and (b'), and we have Our theorems have been obtained in the literature for different special cases, as in Loomis [6] or Zaanen [13].In these cases the tensor products G ⊗ H are already Riesz spaces, which allows a direct construction of the product integral.Also, in these cases the Riesz spaces under consideration fulfil the Stone condition (f ∈ L entails min(f, 1) ∈ L), implying that one may represent the Daniell integrals by Lebesgue integrals and draw on the classical Fubini theorem, too.In the general case we are mainly faced with the task to bridge the gap from The proofs, given in the next section, rely on two results from the literature.Theorem 1 hinges on a density result, concerning the inclusion Fremlin obtained this remarkable result in much greater generality [4,Theorem 4.2 (iii)].His proof is extensive and uses advanced representation theorems for Riesz spaces.A more intrinsic derivation was given by Grobler and Labuschagne [5], still their approach is somewhat involved and demanding.From these considerations only a smaller part is needed in our case, which we present in the subsequent selfcontained Proposition 1 (with a view to readers with little experience in the theory of Riesz spaces).This density statement allows to extent the product measure I from G 1 ⊗ H 1 to the enclosing Riesz space.
We note that the first half of Theorem 1 may as well be derived with marginal efforts from [4, Theorem 5.3].There, Fremlin constructs a linear functional I on the Riesz tensor product L of G and H by approximating I(f ), f ∈ L, from above and from below by the terms I(u) with u ∈ G ⊗ H and u ≥ f or u ≤ f , respectively.We shall use iterated integrals for this purpose.
Theorem 2 is a direct consequence of the second part of Theorem 1 together with the Fubini-Stone theorem [11], which found its way into text books like [12,13].Below, we recall it in Proposition 2 (see also [9,8,3]).

Proofs
The following proposition is a special case of Fremlin [4, Theorem 4.2].Freudenthal's spectral theorem [7, Theorem 40.2] could be used here, too.Our comparatively elementary proof borrows from Grobler et al [5] and is much in the spirit of the spectral theorem.
Recall that a Riesz space L is called Dedekind σ-complete, if it has the property Proposition 1.Let G and H be Dedekind σ-complete Riesz spaces over X and Y .Then for every f ∈ L(G ⊗ H) there is a pair (g, h) ∈ G + × H + and for every ε > 0 some Proof.Preliminary, we recall a well-known property of a Dedekind σ-complete Riesz space L over Z: Let f, u 1 , . . ., u k , v 1 , . . ., v k ∈ L, then as well.Here 1 u<v denotes the characteristic function of the set {z ∈ Z : u(z) < v(z)}.For the proof it is sufficient to consider the case f ≥ 0. Then we have The major task now is to show that L g,h is a Riesz space.It is easy to see that L g,h is a vector space, thus it remains to show that with f ∈ L g,h we also have |f | ∈ L g,h .
First, we consider the case f = with integers a 1 , . . ., a k ∈ {−n, . . ., n}.Apparently, for c > 0 large enough these sets together with the set {g = 0} yield a partition of the domain X.Similar for large c > 0 the sets . ., n} and the set {h = 0} make a partition of Y .
From (1) we obtain that g • 1 Aa 1 ...a k n belongs for all a 1 , . . ., a k to G, and Since the sets A a 1 ...a k n × B b 1 ...b k n are pairwise disjoint, there will be for each f n (x, y) with x ∈ X and y ∈ Y at most one non-vanishing summand in the previous sum.Hence Therefore, in view of This estimate implies that |f | ∈ L g,h .Now let f be any element of L g,h .Then, for any ε > 0 there is a Thus |f | ∈ L g,h .Altogether we have shown that L g,h is a Riesz space.
In order to finish the proof, let L ′ := L g,h , where the union is taken over all (g, h) Proof of Theorem 1. First, we prove that J(K(f )) and K(J(f )) are welldefined for f ∈ L := L(G 1 ⊗ H 1 ).From dominated convergence it follows that the Riesz spaces G 1 and H 1 are Dedekind σ-complete.Thus, for any f ∈ L there are due to Proposition 1 g ⊗ h for some non-negative g ∈ G 1 and h ∈ H 1 .Hence, for any x ∈ X we have Taking the limit n → ∞ and using dominated convergence yields Again taking the limit, dominated convergence yields K(f ) ∈ G 1 .Similarly, we have f y ∈ G 1 and J(f ) ∈ H 1 , hence, we obtain the properties (a) and (b), and we may define a functional I : L → R by for all f ∈ L. It is immediate that I is an integral and that the equation I(g ⊗ h) = J(g)K(h) is fulfilled.Thus we proved existence of a product integral.Now let I ′ : L → R be another integral fulfilling I ′ (g ⊗ h) = J(g)K(h).Then we have Taking the limit n → ∞ yields I(f ) = I ′ (f ), thus we achieve uniqueness of the product measure.In particular, setting I ′ (f ) = K(J(f )) we obtain equality of the two iterated integrals, which finishes the proof.
For the proof of Theorem 2 we resort to Stone's Fubini theorem.Stone [11] uses here the following strengthened version of the conditions (a) and (b): For the proof see [11] or [12, Theorem 7-2.I].
Proof of Theorem 2. Set G = G 1 , H = H 1 , L = L(G 1 ⊗ H 1 ) and I = J ⊗ K in Proposition 2.Then, by Theorem 1 the assumptions of the proposition are met.Thus, by its conclusion it follows I(f ) = J(K(f )) for all f ∈ L 1 .By symmetry, we also obtain I(f ) = K(J(f )).
Remark.We note that with (a") and (b") instead of (a) and (b) the conclusion of the second part of Theorem 1 may fail, as seen from the following example: Let G = H be the Riesz space of all continuous piecewise linear functions on the interval [0, 1].Then the function f (x, y) := y − x belongs to G ⊗ H and f + (x, y) = (y − x) + is an element of L(G ⊗ H).Moreover, setting J(g) = 1 0 g(x) dx for g ∈ G, we obtain an integral J on G. Then J(f + )(y) = 1 0 (y − x) + dx = y 2 /2, thus J(f + ) does not belong to H.
Now the Fubini-Stone theorem reads as follows.Proposition 2. Let G ⊂ F (X), H ⊂ F (Y ) and L ⊂ F (X × Y ) be Riesz spaces and J : G → R, K : H → R and I : L → R integrals.Then, if all f ∈ L fulfil (a"), (b") and the equation I(f ) = J(K(f )), then all f ∈ L 1 fulfil (a'), (b') and I(f ) = J(K(f )).