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Let $(X,\mathcal{F},\mathbf{P})$ be a probability space and $f \colon X \mapsto \mathbf{R}^n$ an integrable function. We assume that $f$ takes its values in a closed convex set $C$ of $\mathbf{R}^n$ and that

$$\int_X f \textrm{d}\mathbf{P} = x$$ is an extremal point of $C$.

Does $f$ coincide with the constant function $x$ ?

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2 Answers

up vote 3 down vote accepted

Yes, $f=x$ almost everywhere wrt $P$. Consider the push-forward measure $\mu=f_*P$ (in probability, it is also called the distribution of $f$). It is defined as $\mu(E)=P(f^{-1}(E))$. This $\mu$ is a probability measure on $C$. Your condition becomes $$\int_C d\mu=x,\quad (1)$$ and this implies that $\mu$ is an atom at $x$.

EDIT. This is a special case of a theorem of Bauer (see, for example Phelps, Lectures on Choquet's theorem, Proposition 1.4). Let $X$ be a non-empty compact convex set in a locally convex space, and $x\in X$. Then $x$ is an extreme point if and only if the point mass at $x$ is the unique measure for which (1) holds. Of course, you did not say that $C$ is bounded, but the reduction to this case is easy.

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This doesn't really answer the question, it just rephrases it. Why does this condition imply $\mu$ is an atom? –  Nate Eldredge Nov 26 '12 at 15:21
    
Thanks for your answer. –  Vincent Beck Nov 26 '12 at 21:19
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Yes (if you mean a.e. coincidence, of course). This can be proved by induction over the dimension. The case $n=1$ is clear, and for the induction step find a closed half-space $H=\lbrace y\in\mathbb R^n: u(y)\le t\rbrace$ for a linear functional $u$ such that $C\subseteq H$ and $u(x)=\alpha$. The one-dimensional case then implies that $u\circ f = t$ a.e., that is, $f$ takes values in the $n-1$-dimensional space $\lbrace u=t \rbrace$.

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Is this fact still true in infinite-dimensional spaces? Say, a separable Frechet space. –  Nate Eldredge Nov 26 '12 at 15:23
    
Yes, if $C$ is compact. See the end of my solution. Bauer's theorem applies in any locally convex space. –  Alexandre Eremenko Nov 26 '12 at 17:39
    
Actually we do not need induction: since $x$ is an extremal point there is a linear functional $u$ such that $u(y) < u(x):=t$ for all $y\in C\setminus\{x\}$ (this is a characterization); with the same argument one immediately has $f(y)\in C\cap\{u=t \}=\{x\}$ a.e. that is $f(y)=t$ a.e. The same argument works with any compact $C$ in a LCTVS, and w.r.to the Pettis integral. –  Pietro Majer Nov 26 '12 at 21:54
    
Okay, one one does not need induction. However, the inductive argument does not need that $C$ is closed. For example, it proves that $\int f d\mu$ always belongs to $C$. –  Jochen Wengenroth Nov 27 '12 at 7:38
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