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This is probably an elementary question, but outside my area of expertise, and I was unable to find any suitable reference:

Suppose $f:X\to E$ is a continuous function from a compact spaces (endowed with a probability Borel measure $\mu$) to a locally convex, Hausdorff, complete topological vector space. We want to define a reasonable integral, i.e., $\int_Xfd\mu$. The question is : when exactly is that one needs weak integrals (e.g., the Gelfand-Pettis integral)? There is a strong integral called the Bochner integral which is typically defined for $E$ a Banach space; but it seems to me that its definition works at least for Frechet spaces. For a general locally convex, Hausdorff, complete TVS, when exactly does Bochner's approach, i.e., to define a strong integral by approximation of $f$ with simple functions (i.e., finite-valued measurable functions) fail?

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A locally convex Hausdorff TVS is not good enough; you need some completeness. Otherwise, you may be able to approximate $f$ by simple functions $f_n$, but find that $\int f_n d\mu$ does not converge. For simple examples, take your measure space to be $\mathbb{N}$ with counting measure. – Nate Eldredge Nov 29 '10 at 23:58
Quasi-completeness (bounded Cauchy nets converge) is a sufficient completeness condition for existence of Gelfand-Pettis integrals. Note that weak* duals of Hilbert spaces are not complete, but are quasi-complete (the issue being the lack of a countable basis at 0 for the topology). The space of test functions is not complete (it is a countable union of nowhere dense closed subsets, which contradicts the conclusion of Baire-Category), but is quasi-complete, as is the space of distributions. @Nate: you might want to refine your example as the measure is required to be finite. – B R Nov 30 '10 at 2:12
The space of test functions is complete because it is LF (see Treves's book "Topological vector spaces", Theorem 13.1) but apparently not metrizable. My question is whether or not in the case of complete spaces one always has a strong integral. – Hadi Nov 30 '10 at 19:17
The Gelfand-Pettis integral exists under the condition that the closure of the convex hull of a compact set be compact. See Paul Garret's paper on this. If you look at the Bochner construction in this case, it works for locally convex spaces, in which the closure of the convex hull of a compact set is complete. So, you have to invest local convexity, but then you exchange the a compactness condition by a completeness, which is (seemingly?) weaker. – Anton Dec 4 '10 at 14:17
Could it be that the issue is not the construction of the Bochner integral but a lack of (strongly) integrable functions? – B R Dec 4 '10 at 18:27


One issue with Bochner integration is that it does not include Riemann-integration. There are Banach-space-valued R-integrable functions that are not B-integrable (example: Consider $X:=\mathcal{l}^p([0,1])$ with $2\leq p < \infty$ and $f:[0,1]\to X, f(t):=e_t$ where $e_t$ is the tupel with exactly one equal to 1 and all other components equal to 0.). The Gelfand-Pettis-Integral on the other hand includes both the Bochner- and the Riemann-Integral.

One problem with B-integration is that you need functions that are almost separable valued (meaning: there is a nullset whose complement has separable image) in order to approximate them with simple functions and this may be a strong restriction. Another issue is that for certain applications the weak topologies just behave better than the strong ones so that Pettis-integration is the natural notion of integration in this cases.

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In which respect is the weak topology better behaved that strong topology? Since strong continuity implies weak continuity, strong should be better, don't you think? – Anton Dec 6 '10 at 7:57
Only if you have strong continuity. If some mapping fails to be strongly continuous one can still hope that it is weak continuous. Also the weak topologies tend to have more compact subsets. – Johannes Hahn Dec 6 '10 at 12:16
Yes sure, but my claim is, that in all relevant applications Bochner's method works, so you indeed have strong topology, which clearly is better. To contradict me, you have to give me an example of a (serious) application where Bochner's appraoach doesn't work. Cheers, Anton – Anton Dec 6 '10 at 12:47
I don't see why I "have to"... but what the hell. Is this a "reasonable application" ? Take a non separable compact hausdorff space and embedded it in any complete LCTVS. This embedding if continuous and compactly supported but can't be Bochner-integrable since it has non-separable image. If you throw out nullsets this depends of course on the measure, but one can choose a space and a probability measure with non-separable non-nullsets. – Johannes Hahn Dec 7 '10 at 11:27
It is indeed Bochner integrable, see below! – Anton Dec 7 '10 at 17:28

In the setting you name, the Bochner approach never fails. As the image $f(X)$ is compact, you construct a net of simple functions as follows. For index set, you take the set of all continuous seminorms $p$ on $E$. For each $p$, let $U_p$ be the open unit-ball defined by $p$. For $n\ge 1$ large enough, you cover $f(X)$ by finitely many translates on $\frac1n U_p$. Make the covering disjoint and make the vectors you translated by the values of a simple function $s_p$ which satisfies $\int_Xp(f-s_p)\,d\mu<1$. The $(s_p)$ form a Cauchy-net, the limit of which is your desired integral.

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The Bochner-Integral is not defined via arbitrary nets, but only via sequences. Otherwise most of the theorie breaks down. For example a limit of a net of measurable functions need not to be measurable etc. – Johannes Hahn Dec 7 '10 at 11:22
The Bochner integral can easily bedefined for arbitrary nets. You simply make measurability part of the definition. So you say a measurable function is Bochner-approximable, if there exists a net $s_j$ of simple functions such that $\int_Xp(f-s_j)$ tends to zero for every continuous seminorm. In that case the net $\int_Xs_j$ is Cauchy, so, if the space is complete, you are done. If the function is bounded, quasi-completeness is enough. – Anton Dec 7 '10 at 16:22
This isn't the definition of the Bochner integral so you're talking about another notion of integration. – Johannes Hahn Dec 7 '10 at 20:23
Well, that was the question, wasn't it? – Anton Dec 7 '10 at 21:29
Just a question beside: What if the preimage of the translates have infinite measure? – Freeze_S May 19 '14 at 19:36

Anton's answer prompted me to pull out Rudin's Functional Analysis. Exercise 23 of Chapter 3 is to exhibit the Gelfand-Pettis integral of a continuous function with values in a Frechet space as a strong limit of "Riemann sums" (looks slightly stronger than the Bochner integral, in fact). I couldn't make sense of how to use the hint, so I decided to see what I could come up with. It's been a long time since I've done this sort of thing, so I may have done something stupid.

We're going to assume $X$ is compact, $\mu$ is a probability Borel measure, $f$ is continuous, $E$ is locally convex and satisfies the technical criterion for the existence of Gelfand-Pettis integrals (closure of the convex hull of the image is compact). Then $f(X)$ is compact. If we also assume either $E$ is a metric space or $X$ is separable, then $f(X)$ is separable. Take a countable dense subset $e_i$ of $f(X)$. For a finite subset $B$ of seminorms and $\epsilon>0$, let $V(B,\epsilon,e_i)=\{e:p(e-e_i)<\epsilon\}$. These form a basis for the topology of $E$ (and they are convex and balanced).

First, we're going to construct a sequence of simple functions that converges to $f$, mimicking the example from measure theory.

Since $f(X)$ is compact, we can form a cover with finitely many of the $V(B,\epsilon,e_i)$, call them $A_i$. Let $E_i=A_i-\bigcup_{j=1}^{i-1}A_j$. Set $X_i=f^{-1}(E_i)$, so the $X_i$ are a disjoint partition of $X$.

Set $g_{B,\epsilon}(x)=\sum_i 1_{X_i}(x)e_i$ (where $i$ ranges over a finite set). Consider $f(x)-g_{B,\epsilon}(x)$. If $x\in X_i$, $f(x)\in E_i$, so for all $p\in B$, $p\big(f(x)-g_{B,\epsilon}(x)\big)=p\big(f(x)-e_i\big)<\epsilon$. Convergence is easy from this.

Now we will show that $\int_X g_{B,\epsilon}\ d\mu=\sum_i \mu(X_i)e_i$ converges. We "cheat" by showing it converges to the Gelfand-Pettis integral of $f$ (instead of e.g. showing that it is Cauchy). So consider $$\int_X f\ d\mu- \int_X g_{B,\epsilon}\ d\mu=\int_X f- g_{B,\epsilon}\ d\mu=\sum_i\int_{E_i}f-e_i\ d\mu$$

On $E_i$, $f-e_i$ lies in $V(B,\epsilon)$, so $\int_{E_i}f-e_i\ d\mu$ is in the closure of $\mu(E_i)\cdot V(B,\epsilon)\subset \overline{V(B,\epsilon)}$ (since $V(B,\epsilon)$ is balanced) (this uses the estimate that the Gelfand-Pettis integral lies in the closure of the convex hull of the image, times the measure of the space). So $\sum_i\int_{E_i}f-e_i\ d\mu\subset \sum_i \overline{V(B,\epsilon)}\subset \overline{\sum_i V(B,\epsilon)}$.

By some basic tvs theorems, since $V(B,\epsilon)$ forms a basis for the topology, we can choose $V(B,\epsilon)$ so that $\overline{\sum_i V(B,\epsilon)}$ is inside any given neighborhood of $0$ (see p.10-11 of Rudin).

So Bochner integrals of continuous functions exist whenever Gelfand-Pettis integrals exist (plus a separability assumption), unless I made a mistake. Even though I left it in, I don't think that I used separability in an essential way (I could have just chosen an uncountable dense subset $e_i$ and proceeded). That requirement for strong measurability (almost-separably valued) may be an artifact of using sequences instead of nets.

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