Let $X$ be a compact metric space, and let $\mathcal P(X)$ be the (compact, metrisable) space of Borel probability measures on $X$. Similarly, $\mathcal P (\mathcal P (X))$ is the space of Borel probability measures on $\mathcal P (X)$. I want to make sense of the integral $$ \int_{\mathcal P(X)} \mu \,\mathrm d \mathcal \tau (\mu) \tag{INT}\label{INT} $$ where $\tau \in \mathcal P (\mathcal P (X))$. Is there an elementary definition of this integral? (See the end of the question for an elaboration of what I mean by "elementary".)

It seems one way to interpret \eqref{INT} is to view it as a Bochner integral, by considering $\mathcal P(X)$ as a (compact, convex) subset of the Banach space of finite signed Borel measures on $X$ with the total variation norm. One property of the Bochner integral is that, for any bounded operator $T\colon \mathcal P(X) \to Y$ where $Y$ is another Banach space, we have that $$ \int_{\mathcal P (X)} T\mu\,\mathrm d\tau (\mu) = T \left( \int_{\mathcal P(X)} \mu \,\mathrm d \mathcal \tau (\mu) \right). \tag{*}\label{*} $$

In particular, suppose that $Y = L(C(X),\mathbb R)$, the space of bounded linear functionals on $C(X)$, and $T$ is the operator given by $$ \mu \mapsto \left(\begin{align*} C(X) &\to \mathbb R \\ f &\mapsto \int_X f(x) \,\mathrm d \mu(x) \end{align*}\right). $$ Then, we can parse \eqref{*} as $$ \int_X f(x) \, \mathrm d \mu_0 (x) = \int_{\mathcal P(X)} \int_X f(x) \,\mathrm d \mu(x) \,\mathrm d \tau (\mu) \tag{+} \label{+} $$ for each $f \in C(X)$, where $\mu_0$ is just the value of \eqref{INT}.

Hopefully, everything I've said so far is correct. (Please do point out mistakes if there are any.) If so, can I take \eqref{+} as the definition of \eqref{INT}? That is, I would like to define \eqref{INT} as the $\mu_0 \in \mathcal P(X)$ that satisfies \eqref{+} for each $f \in C(X)$. Do I lose any important properties of the integral by using this definition?

I am looking for a definition of \eqref{INT} without too much additional machinery, beyond what you would see in a typical introductory measure theory or functional analysis course. I was hoping \eqref{+} would do, but I am also open to alternative suggestions.

Additionally, if anyone can recommend a nice, readable reference for this material, I'd appreciate that as well. An ideal reference would be one that's accessible to someone who only knows a little measure theory and functional analysis, but I'll take what I can get.

  • $\begingroup$ I'm not sure if this is what you're looking for, but you do not need a lot of machinery to set up the Bochner integral. These lecture notes of Marc Rieffel (which define Bochner integration in Chapter 4 -- everything before that is standard measure theory) assume nothing but the definition of a Banach space and some point-set topology: math.berkeley.edu/~rieffel/measinteg.html $\endgroup$ May 11, 2020 at 16:51
  • $\begingroup$ @AidanBackus Thanks for the notes; I'll check them out. I agree that the Bochner integral does not require a lot of machinery; in fact, (as far as I understand) it is the most natural extension of the Lebesgue integral to Banach spaces. However, this question is motivated by research I'm doing in economics, and I think the audience I'm writing for there might take better to definitions that use objects they're already likely to know something about (e.g. standard Lebesgue integrals). $\endgroup$ May 11, 2020 at 17:26
  • $\begingroup$ That's a fair point; I tried to think of a nice way to reduce the Bochner integral to the Lebesgue integral in general but I don't know how to do it without appealing to the Hanh-Banach theorem or something else that economists might not be comfortable with. $\endgroup$ May 11, 2020 at 17:31
  • $\begingroup$ @AidanBackus That seems interesting! Please feel free to include it in your answer. $\endgroup$ May 11, 2020 at 17:33
  • $\begingroup$ Sure, I outlined the argument. The hard part if you approach it this way is showing that the Bochner integral actually exists, but for an economics talk that could probably just be blackboxed :-) $\endgroup$ May 11, 2020 at 17:47

2 Answers 2


This is a really interesting question! I think $(+)$ is the only sensible definition of $\mu_0$. As you correctly observed, this follows once we've decided to define $\mu_0$ using Bochner integration.

But the Bochner integral is the unique linear map satisfying a few "obvious" criteria, like the $L^1$ triangle inequality and dominated convergence. See Bochner Integral: Axioms. If you want to define $\mu_0$ in any way that is vaguely useful, it had better satisfy those axioms.

Per the discussion in the comments, let me outline how one could express the Bochner integral in terms of the Lebesgue integral. Let $X$ be a measure space, $B$ a Banach space, $B^*$ its dual, and $L^1(X \to B)$ the space of Bochner-integrable functions $X \to B$.

By the Hanh-Banach theorem, for every $x,y \in B$, $x=y$ if and only if for every $\varphi \in B^*$, $\varphi(x) = \varphi(y)$. In other words, to specify a point $x \in B$, it suffices to specify $\varphi(x)$ for every linear functional $\varphi$ (though there is no guarantee that $x$ will exist, and this seems unfixable due to Enflo's theorem that there are Banach spaces which do not admit a Schauder basis, if $x$ exists this uniquely defines it; maybe $\mathcal P(X)$ actually has a Schauder basis, and then you can just use the Schauder basis rather than fiddle around with linear functionals like this).

As you observe, the Bochner integral commutes with bounded linear maps. In particular it commutes with bounded linear functionals, i.e. elements of $B^*$. So if $f \in L^1(X \to B)$, we have $$\varphi\left(\int_X f(x)~dx\right) = \int_X \varphi(f(x)) ~dx.$$ Here the integral $\int_X \varphi(f(x))~dx$ is just the plain jane Lebesgue integral. We have specified $\varphi\int_Xf$ for every $\varphi$, so we have specified $\int_Xf$, provided that $\int_Xf$ actually exists. (It does, by the usual construction of the Bochner integral in terms of ISF, but the point of this outline was to give a cheap definition that averts this modulo technical details.)

  • $\begingroup$ Thanks for that! The argument you outline is roughly the idea behind Simon's proof of the theorem I state in my answer. He just uses a nice trick to keep track of all the $\phi$'s at the same time, and then leverages compactness. I am confused about one thing, though: do we really need Hahn-Banach to conclude that $B^*$ separates points? I can see how it follows from HB, but that seems like an extremely overpowered way to do it. $\endgroup$ May 12, 2020 at 16:34
  • $\begingroup$ In the case you're interested in I think it can be avoided. Here B* is the double-dual of C(X), and I think C(X) should already separate points in P(X). $\endgroup$ May 12, 2020 at 17:37
  • $\begingroup$ In general it cannot be avoided. The reason is that in certain models of set theory without the axiom of choice, there exist Banach spaces B such that B* is the trivial vector space! So we must use the axiom of choice somehow, and this manifests as the Hanh-Banach theorem. $\endgroup$ May 12, 2020 at 17:39
  • $\begingroup$ Yes, the general case was what I was interested in, but I was happy to keep AC. The argument I had in mind was to construct separating linear functionals using the basis on $B$, but of course, the existence of such a basis depends on AC. This strikes me as simpler than relying on Hahn-Banach, but I suppose that is a matter of taste. $\endgroup$ May 12, 2020 at 17:41
  • $\begingroup$ I think any argument that uses Hanh-Banach can be replaced by using AC to choose a basis and induct on its (transfinite) length, but yeah that does come down to taste. $\endgroup$ May 12, 2020 at 17:42

I think I have an answer, but there are some details I need to think about more carefully.

Yes, we can take (+) as the definition of (INT). This follows from a version of the Strong Krein-Milman Theorem, which I state below. (This is Theorem 9.1 in Simon; reference below.) We denote the continuous dual of a topological vector space $X$ by $X^*$, and the closed convex hull of a set $A$ by $\mathrm{cch}(A)$.

Let $A$ be a compact convex subset of a real locally convex (Hausdorff) vector space $X$ and let $\mu \in \mathcal P(A)$. Then, there is a unique point $r(\mu)\in A$, called the barycenter or resultant of $\mu$, so that for any $\ell \in X^*$, $$ \ell (r (\mu)) = \int_A \ell(x)\,\mathrm d \mu(x). $$ The map $r$ is a continuous affine map of $\mathcal P(A)$ (with the weak-$*$ topology) onto A and is the unique such map with $r(\delta_x) = x$. More generally, if $B \subset A$ is closed, and $\nu (A \setminus B) = 0$, then $r(\nu) \in \mathrm{cch}(B)$ and $$ r[\{ \nu : \nu(A\setminus B) = 0 \}] = \mathrm{cch}(B). $$

This means that the measure $\mu_0$ defined by (+) is well-defined, and guarantees that the integral (INT) is continuous in $\tau$.


Simon, B. (2011). Convexity: an analytic viewpoint (Vol. 187). Cambridge University Press.


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