# Convex representation of a measure

Let $\mathcal P(X)$ denote the space of all probability measure defined on a measurable space $X$. We canonically endow the former with its own measurability structure, generated by evaluation maps. Let $P \subseteq \mathcal P([0,1])$ be a measurable subset of probability measures, and let $\hat p\notin P$ be such that for every bounded measurable $f:[0,1] \to \Bbb R$ there exists $p_f\in P$ satisfying $$\int_{[0,1]} f(x) \hat p(\mathrm dx) = \int_{[0,1]} f(x) p_f(\mathrm dx),$$ or $\hat p f = p_f f$ in a short form. Does it necessarily means that there exists a probability measure $\nu \in \mathcal P(\mathcal P([0,1]))$ such that $\nu(P) = 1$ and $\hat p = \int_P p\,\nu(\mathrm dp)$?

I think this result is quite easy to show for finite $X$ where $\mathcal P(X)$ is just a subset of $\Bbb R^n$, however I am not sure whether it still holds true in my more general case. Obviously, from $[0,1]$ it would generalize to any Borel space.

• In case this is a standard borel space this could be seen using hahn-banach by taking a compact model. Sep 26, 2016 at 8:58
• @UriBader: can you be more specific, please? It is indeed a standard Borel space
– SBF
Sep 26, 2016 at 9:06
• Sorry, my previous remark is incorrect. Take P to be all non-delta mearues. Sep 26, 2016 at 9:07
• I should go on flight mode... I will visit here in 14hrs. Sep 26, 2016 at 9:10
• @UriBader: thanks, have a nice flight. Having $P$ being all non-delta measures does not provide a counterexample to the OP though (if that's what you've meant). Let's say $\hat p = \delta(0)$, then taking $f = 1_{\{0\}}$ means that $\hat p f = 1$ but $p f = 0$ for all $p\in P$.
– SBF
Sep 26, 2016 at 9:17

I understand the condition on $$p_f$$ as belonging to the closure $$\bar{P}$$ for the weak-$$\star$$ topologie. And one can then ask whether $$[\int p d\nu(p),\nu\in \mathcal{P}(\mathcal{P}([0,1]))]$$ contain this closure. Consider the following counter example $$P=[\lambda \delta_0+(1-\lambda)\delta_1:0<\lambda<1]$$ Then $$\delta_0 \in \bar{P}$$ but not to in the convex set.

Here we have $$p_ff=\hat{p}f$$ which is a stronger condition than $$\inf_{p\in P}|pf-\hat{p}f|=0$$. We therefore have to work a bit harder to construct the counter example.

Let $$\hat{p}=\frac{1}{2}\delta_0+\frac{1}{2}1_{[0,1]}$$ and choose $$P=\{q_x :x\geq \frac{1}{2}\}\cup[p\in \mathcal{P}([0,1]):p1_{\{0\}}<\frac{1}{2}]$$ with $$q_x=\frac{1}{2}\delta_0+\frac{1}{2}\delta_x$$ We see that if $$\hat{p}=\int_P pd\nu(p)$$ then $$\hat{p}1_{\{0\}}=\frac{1}{2}=\frac{1}{2}\nu(\{q_x : x\geq \frac{1}{2}\})+\int_{P-\{q\}} p1_{\{0\}}d\nu(p)$$ and therefore the support of $$\nu$$ is a subset of $$\{q_x :x\geq \frac{1}{2}\}$$ which is impossible.

We now check that $$P$$ satisfies the condition. Let $$f$$ a bounded (measurable) function.

If $$f(0)=a$$ and $$f(t)=b$$ for all $$t\in E\subset [0,1]$$ a set of Lebesgue measure 1. Then there exist $$x\geq \frac{1}{2}$$ such that $$f(x)=b$$ and we have $$\hat{p}f=\frac{1}{2}(a+b)=q_x f$$.

If $$f$$ is not constant on a set of Lebesgue measure 1. then there exists $$\epsilon >0$$ and $$\tilde{p}$$ with $$\tilde{p}1_{\{0\}}=0$$ and $$\tilde{p}f>(1+\epsilon)\int f(t)dt$$ Suppose $$f(0)\geq \int f(t)dt$$. Then by continuity we can find $$\lambda<\frac{1}{2}$$ such that $$\lambda f(0)+(1-\lambda)\tilde{p}f=\hat{p}f$$.

Suppose $$f(0)\leq \int f(t)dt$$ then we use the same argument but with $$\tilde{p}f<(1-\epsilon)\int f(t)dt$$

• How do you deal with the measurable function that is 1 at 0 and vanishes everywhere else? Oct 6, 2018 at 16:54
• @MichaelGreinecker I think one then chooses $q_x$ for any $x$ which has the same mass as $\hat{p}$ at 0. As I understand it, that is precisely the idea: Since for such functions one would have to choose a $q_x$, there is no convex representation of $\hat{p}$ that fits all functions. Overall I think the proof is correct, even though at the end the case $f(0) = \int f(t) dt$ should perhaps be treated separately as $\tilde{p} f = \int f(t) dt$ must be chosen. Oct 7, 2018 at 20:49
• Thanks, not sure I understand the notation here. What is a $1_{[0,1]}$ in definition of $\hat p$, a Lebesgue measure? You also a similar notation below $1_{\{0\}}$, where I guess it means the indicator function instead.
– SBF
Oct 11, 2018 at 15:06
• @Ilya . Yes, $1_{[0,1]}$ is just the Lebesgue measure with density the function $f(x)=1$ for all $x\in [0,1]$ Oct 11, 2018 at 15:27
• I'm not familiar with equating measures to functions. Do you mean that $\hat p$ is half a Lebesgue measure on $[0,1]$ with another half-mass at $0$?
– SBF
Oct 11, 2018 at 15:30