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 $A_\delta$.)

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).