The question is: Consider the set $X$ with measure $\sigma(X)=1$. Let $K$ be a compact convex subset of $\mathbb{C}$. Let $f$ be a complex valued, measurable function defined on $X$ such that the range of $f$ is in $K$. Show that $f$ is integrable w.r.t. $\sigma$ and that $$ \int_X f(x)\sigma(dx) \in K $$ What have I done so far: I was able to show that $f$ is integrable w.r.t. to $\sigma$. However, I am having a lot of grief showing that the integral is in the compact convex set. I can see that this is true for a small non-trivial case like say a unit square with the four corners having the value $\frac14$ but this is proving to be beyond me. I have consulted Rudin's Real and Complex Analysis and Folland's Real Analysis but to no avail. It seems like this might be a theorem of a corollary of a theorem and I am missing a link. Any suggestions?

  • $\begingroup$ Anyone? Anything? $\endgroup$ – jbond Dec 15 '15 at 2:37

An informal argument is that, since the values of $f$ are in $K$ a compact convex subset of $\mathbb{C}$ and $\int_X f(x)\sigma (dx)$ is an average of the values of $f$, then this average exists (since $f$ is finite) and must belong to $K$.

Now I will show a formal proof. In this proof, I will assume you are familiar with Hahn-Banach separation theorem I

By contradiction, assume $z_0=\int_X f(x)\sigma (dx)\notin K$. Using $K$ and $\{z_0\}$ are compact convex disjoint sets, by Hahn-Banach separation theorem, there exists $v\in \mathbb{C}$ and $d\in\mathbb{R}$, s.t. $$ \langle v,z\rangle < d < \langle v,z_0\rangle $$ for all $z\in K$. Then $$ \langle v,f(x)\rangle < d < \langle v,z_0\rangle $$ for all $x\in X$.

Let $g:X\to \mathbb{R}$ be defined by $g(x)=\langle v,f(x)\rangle$. We can show $g$ is a measurable and integrable function, using $f$ measurable and integrable.

On the one hand, we can show \begin{eqnarray} \int_X g(x)\sigma (dx) dx & = &\int_X \langle v,f(x)\rangle\sigma(dx)\\ & = & \langle v,\int_X f(x)\sigma(dx)\rangle\\ & = & \langle v,z_0\rangle \end{eqnarray} while on the other hand, $$ \langle v,z_0\rangle=\int_X g(x)\sigma (dx) dx<d<\langle v,z_0\rangle $$ which is a contradiction.

  • $\begingroup$ Aha! The Hahn-Banach Sep Thm 1 is the key. Thank you so, so much! $\endgroup$ – jbond Dec 15 '15 at 18:28
  • $\begingroup$ You are welcome. Glad you found my answer useful. $\endgroup$ – Nate River Dec 15 '15 at 23:50

Another approach can based on the distance-to-the-boundary function. As a bonus, $K$ need only be convex and bounded. (For unbounded $K$, one must assume that $\int|f(x)|\,\sigma(dx)<\infty$.) Define $\varphi(z)=\inf\{|z-w|:w\in\partial K\}$ (the distance to the boundary $\partial K$ of $K$) for $z\in K$ and $\varphi(z)=-\infty$ for $z\notin K$. Then $\varphi:\Bbb C\to[-\infty,\infty)$ is concave. Let $m$ denote $\int_X f(x)\,\sigma(dx)$. By the vector-valued form of Jensen's inequality,
$$ \varphi(m)\ge\int_X \varphi(f(x))\,\sigma(dx)\ge 0, $$ the final inequality following because $f$ takes values in $K$ and $\varphi(z)\ge 0$ for all $z\in K$. It follows that $\varphi(m)\ge 0$; in particular, $\varphi(m)>-\infty$, so $m\in K$.

This argument works in all dimensions, and more generally in topological vector spaces for which (i) barycenters of probability measures exist and (ii) an appropriate form of Jensen's inequality holds.

  • $\begingroup$ Thanks for the alternative route, John. $\endgroup$ – jbond Dec 16 '15 at 2:40
  • $\begingroup$ This is a perfect example of a flawed proof. Jensen's inequality is generally stated for proper, that is real-valued, convex functions. If you want to extend it to convex functions $\varphi$ taking values in $(-\infty,\infty]$, then it amounts to prove that if $\int \varphi \circ f d\mu$ is finite (so that f takes a.e. values in the effective domain of $\varphi$), then $\int f d \mu$ is in the effective domain of $\varphi$, which is what we want to prove! PS For the definition of effective domain of a convex function, see e.g. Rockafellar, Convex Analysis. $\endgroup$ – Maurizio Barbato Sep 2 '17 at 13:41
  • $\begingroup$ Nate River's proof works well, and for another proof you can see e.g. Dudley, Real Analysis and Probability, Theorem (10.2.6). $\endgroup$ – Maurizio Barbato Sep 2 '17 at 13:43
  • $\begingroup$ Anyhow, I want to thank John Dawkins for making me know this wonderful proof: the distance function $f(x)=d(x,\partial C)$ from the boundary of an open convex set $C \subset \mathbb{R}^n$ is a concave function on $C$. For a proof, see e.g. Güler, Foundations of Optimization, Theorem (6.20) or Hörmander, Notions of Convexity, Theorem (2.1.24). In the case $n=2$, you can prove the Theorem in a pure geometric way. Take two points $x, y \in C$, and draw the open discs $D_x$ and $D_y$ with centers $x$ and $y$ and radii $f(x)$ and $f(y)$ respectively. $\endgroup$ – Maurizio Barbato Sep 2 '17 at 13:49
  • $\begingroup$ Then, for every $t \in (0,1)$ the open disc with center $tx+(1-t)y$ and radius $tf(x) +(1-t)f(y)$ is contained in the convex hull of $D_x$ and $D_y$, and so in $C$. QED $\endgroup$ – Maurizio Barbato Sep 2 '17 at 13:50

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