Commuting supremum and expectation

Question

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.

Answer 1

Clearly, $M(\omega) = \sup_{a\in U} g(a,S_t)$ is $\mathcal F_t$-measurable.

Define for $\delta>0$ $$ \mathfrak A_\delta = \{(a,\omega)\in U\times \Omega\mid g(a,\omega)>M(\omega)-\delta\} $$ 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 $\mathcal F_t$-measurable $A_\delta$ such that $(A_\delta(\omega),\omega)\in \mathfrak A_\delta$ almost surely. Hence $E[g(A_\delta,S_t)]\ge E[M(\omega)]-\delta$. 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 $A_\delta$.)

UPD: This is perhaps wrong (I just noticed Did's comment).

Answer 2

I think it is quite easy to show

$E\left[\sup\limits_{a\in U}\;g(a,S_t)\right]\leq \sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U} E\left[g(A,S_t)\right]$

Proof:

$S$ is $\mathcal F$-adapted, which implies: $\sigma(S_t)\subseteq\mathcal F_t$

From that it follows,

$\sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U} E\left[g(A,S_t)\right]\geq \sup\limits_{A,\;\sigma(A)\subseteq\sigma(S_t),\;A(\omega)\in U} E\left[g(A,S_t)\right]$

the supremum is taken over all $\sigma(S_t)$-measurable $A$.

We know that, for any $S_t$ there exists a increasing sequence $a^{S_t}_n \in U$ with

$\lim\limits_{n\rightarrow\infty}g(a^{S_t}_n,S_t)=\sup\limits_{a}\;g(a,S_t)$

For every $n$, $a^{S_t}_n$ is clearly a random variable measurable on $\sigma(S_t)$. From that it follows:

$\sup\limits_{A,\;\sigma(A)\subseteq\sigma(S_t),\;A(\omega)\in U} E\left[g(A,S_t)\right]\geq E\left[g(a^{S_t}_n,S_t)\right] \quad\forall n$

i.e.

$ \sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U} E\left[g(A,S_t)\right]\geq E\left[\sup\limits_{a\in U}\;g(a,S_t)\right]$