Given a one-parametric random function on a probability space $(\Omega,\mathcal F,\mathbb P)$:

$X:U\times\Omega\to \mathbb R \text{ and } (a,w)\mapsto X(a,w), \text{ with } \sigma(X(a))\subseteq \mathcal F\quad\quad\forall a\in U\subseteq\mathbb R, $

Then the following holds:

$E\left[\sup\limits_{a\in U}X(a)\right]=\sup\Bigr\lbrace E\left[X(A)\right]\Bigr|\sigma(A)\subseteq{\mathcal F},\;A(\omega)\in U\Bigr\rbrace$ and also $E\left[\sup\limits_{a\in U}X(a)\right]=\sup\Bigr\lbrace E\left[X(A)\right]\Bigr|\sigma(A)\subseteq{\bigcup\limits_{a\in U}\sigma(X(a))},\;A(\omega)\in U\Bigr\rbrace$

## Proof:

The following holds trivially:

$E[X(A)]\le E[\sup_{a\in U} X(a)]$

it remains to show the other direction. This is done by applying zhoraster's answer:

Clearly, $M(\omega) = \sup_{a\in U} X(a,w)$ is $\mathcal F$-measurable.

Define for $\delta>0$

$\mathfrak A_\delta = \lbrace(a,\omega)\in U\times \Omega\mid X(a,w) >M(\omega)-\delta\rbrace$

This set is in $\mathcal B(\mathbb R)\otimes \mathcal F_t$, and it has a full projection onto $\Omega$. By a measurable selection theorem (which I think one can find in Bogachev *Measure Theory*) there is an $\mathcal F$-measurable $A_\delta$ such that $(A_\delta(\omega),\omega)\in\mathfrak A_\delta$ almost surely. Hence $E[X(A_\delta)]≥E[M(\omega)]−\delta$. We get the desired statement by letting $\delta\to 0$.

(One can also use Kuratowski--Ryll-Nardzewski theorem to prove the existence of a measurable $A_\delta$.)

After a very good answer of zhoraster, I realized, that my initial question was a mixup of several different things. Thats why I changed it community wiki and clearified the problem.

clearlyor not. :-) .../... $\endgroup$ – Did Feb 15 '11 at 13:27