Return to Answer

This will require very strong conditions on $G$. The most general result I know of is an "almost upward-filtering" condition, fairly well known in stochastic optimal control theory: Assume $G(t,\cdot)$ is measurable for each $t \in [0,1]$, and $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$. Then $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\text{ess}\sup_{t \in [0,1]}G(t,X)]$ if and only if for all $\epsilon > 0$ and $r,s \in [0,1]$ there exists $t \in [0,1]$ such that $\mathbb{E}[(G(t,X) - G(s,X) \vee G(r,X))^-] \le \epsilon$. I believe this result is originally due to J.A. Yan, in the hard-to-find paper "On the commutability of essential infimum and conditional expectation operators".

Note: $\sup_{t \in [0,1]}G(t,X)$ need not be measurable, which is why the essential supremum is used in the aforementioned theorem.

A more transparent condition can be derived from the above if we add a continuity assumption: Assume $G(\cdot,x)$ is continuous for each $x$, $G(t,\cdot)$ is measurable for each $t$, and $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$. Then $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)]$ if and only if there exists $T \in [0,1]$ such that $G(T,X) = \sup_{t \in [0,1]}G(t,X)$ almost surely.

PROOF: The continuity assumption guarantees that $\sup_{t \in [0,1]}G(t,X)$ is indeed measurable (e.g. by Theorem 18.19 of Aliprantis & Border), and thus $$ \text{ess}\sup_{t \in [0,1]}G(t,X) = \sup_{t \in [0,1]}G(t,X). $$ The aforementioned theorem and a simple argument using compactness of $[0,1]$ and Fatou's lemma shows that (under our assumptions) $$ \sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)] $$ if and only if for all $$ \mbox{$r,s \in [0,1]$ there exists $t \in [0,1]$ such that $G(t,X) \ge G(r,X) \vee G(s,X)$ a.s.} $$ Since the "if" part is trivial, we now prove the "only if". Consider the set $S := \{G(t,X) : t \in [0,1]\}$ with the partial order given by almost sure inequality. Compactness of $[0,1]$ and continuity of $G(\cdot,x)$ for all $x$ yield the existence of an upper bound in $S$ for any chain of $S$, and thus by Zorn's lemma $S$ contains a maximal element. That is, there exists $T \in [0,1]$ such that there is no $s \in [0,1]$ for which $$ \mbox{$G(s,X) \ge G(T,X)$ a.s. and $P(G(s,X) > G(T,X)) > 0$.} $$ For any $t \in [0,1]$ there exists $r \in [0,1]$ such that $$ G(r,X) \ge G(T,X) \vee G(t,X) \ge G(T,X) \mbox{a.s.} \,, $$ which implies $G(r,X) = G(T,X)$ a.s. and thus $G(T,X) \ge G(t,X)$ a.s.. Hence $G(T,X) \ge G(t,X)$ a.s. for any $t \in [0,1]$. $\blacksquare$

This will require very strong conditions on $G$. The most general result I know of is an "almost upward-filtering" condition: Assume $G(t,\cdot)$ is measurable for each $t \in [0,1]$, and $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$; then $$ \sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\text{ess}\sup_{t \in [0,1]}G(t,X)] $$ if and only if for all $\epsilon > 0$ and $r,s \in [0,1]$ there exists $t \in [0,1]$ such that $$ \mathbb{E}[(G(t,X) - G(s,X) \vee G(r,X))^-] \le \epsilon. $$ I believe this result is originally due to J.A. Yan, in the hard-to-find paper "On the commutability of essential infimum and conditional expectation operators" and has since become become fairly well-known in the stochastic control literature.

Note: $\sup_{t \in [0,1]}G(t,X)$ need not be measurable, which is why the essential supremum is used in the aforementioned theorem.

A more transparent condition can be derived from the above if we add a continuity assumption: Assume $G(\cdot,x)$ is continuous for each $x$, $G(t,\cdot)$ is measurable for each $t$, and $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$. Then $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)]$ if and only if there exists $T \in [0,1]$ such that $G(T,X) = \sup_{t \in [0,1]}G(t,X)$ almost surely.

PROOF: The continuity assumption guarantees that $\sup_{t \in [0,1]}G(t,X)$ is indeed measurable (e.g. by Theorem 18.19 of Aliprantis & Border), and thus $$ \text{ess}\sup_{t \in [0,1]}G(t,X) = \sup_{t \in [0,1]}G(t,X). $$ The aforementioned theorem and a simple argument using compactness of $[0,1]$ and Fatou's lemma shows that (under our assumptions) $$ \sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)] $$ if and only if for all $$ \mbox{$r,s \in [0,1]$ there exists $t \in [0,1]$ such that $G(t,X) \ge G(r,X) \vee G(s,X)$ a.s.} $$ Since the "if" part is trivial, we now prove the "only if". Consider the set $S := \{G(t,X) : t \in [0,1]\}$ with the partial order given by almost sure inequality. Compactness of $[0,1]$ and continuity of $G(\cdot,x)$ for all $x$ yield the existence of an upper bound in $S$ for any chain of $S$, and thus by Zorn's lemma $S$ contains a maximal element. That is, there exists $T \in [0,1]$ such that there is no $s \in [0,1]$ for which $$ \mbox{$G(s,X) \ge G(T,X)$ a.s. and $P(G(s,X) > G(T,X)) > 0$.} $$ For any $t \in [0,1]$ there exists $r \in [0,1]$ such that $$ G(r,X) \ge G(T,X) \vee G(t,X) \ge G(T,X) \mbox{a.s.} \,, $$ which implies $G(r,X) = G(T,X)$ a.s. and thus $G(T,X) \ge G(t,X)$ a.s.. Hence $G(T,X) \ge G(t,X)$ a.s. for any $t \in [0,1]$. $\blacksquare$

This will require very strong conditions on $G$. The most general result I know of is an "almost upward-filtering" condition, fairly well known in stochastic optimal control theory: Assume $G(t,\cdot)$ is measurable for each $t \in [0,1]$, and $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$. Then $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\text{ess}\sup_{t \in [0,1]}G(t,X)]$ if and only if for all $\epsilon > 0$ and $r,s \in [0,1]$ there exists $t \in [0,1]$ such that $\mathbb{E}[(G(t,X) - G(s,X) \vee G(r,X))^-] \le \epsilon$. I believe this result is originally due to J.A. Yan, in the hard-to-find paper "On the commutability of essential infimum and conditional expectation operators". Note that $\sup_{t \in [0,1]}G(t,X)$ need not be measurable, which is why the essential supremum is used in the aforementioned theorem.

A more transparent condition can be derived from the above if we add a continuity assumption: Assume $G(\cdot,x)$ is continuous for each $x$, $G(t,\cdot)$ is measurable for each $t$, and $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$. Then $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)]$ if and only if there exists $T \in [0,1]$ such that $G(T,X) = \sup_{t \in [0,1]}G(t,X)$ almost surely.

PROOF: The continuity assumption guarantees that $\sup_{t \in [0,1]}G(t,X)$ is indeed measurable (e.g. by Theorem 18.19 of Aliprantis & Border), and thus $\text{ess}\sup_{t \in [0,1]}G(t,X) = \sup_{t \in [0,1]}G(t,X)$. The aforementioned theorem and a simple argument using compactness of $[0,1]$ and Fatou's lemma shows that (under our assumptions) $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)]$ if and only if for all $r,s \in [0,1]$ there exists $t \in [0,1]$ such that $G(t,X) \ge G(r,X) \vee G(s,X)$ a.s.. Since the "if" part is trivial, we now prove the "only if". Consider the set $S := \{G(t,X) : t \in [0,1]\}$ with the partial order given by almost sure inequality. Compactness of $[0,1]$ and continuity of $G(\cdot,x)$ for all $x$ yield the existence of an upper bound in $S$ for any chain of $S$, and thus by Zorn's lemma $S$ contains a maximal element. That is, there exists $T \in [0,1]$ such that there is no $s \in [0,1]$ for which $G(s,X) \ge G(T,X)$ a.s. and $P(G(s,X) > G(T,X)) > 0$. For any $t \in [0,1]$ there exists $r \in [0,1]$ such that $G(r,X) \ge G(T,X) \vee G(t,X) \ge G(T,X)$ a.s., which implies $G(r,X) = G(T,X)$ a.s. and thus $G(T,X) \ge G(t,X)$ a.s.. Hence $G(T,X) \ge G(t,X)$ a.s. for any $t \in [0,1]$.

This will require very strong conditions on $G$. The most general result I know of is an "almost upward-filtering" condition, fairly well known in stochastic optimal control theory: Assume $G(t,\cdot)$ is measurable for each $t \in [0,1]$, and $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$. Then $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\text{ess}\sup_{t \in [0,1]}G(t,X)]$ if and only if for all $\epsilon > 0$ and $r,s \in [0,1]$ there exists $t \in [0,1]$ such that $\mathbb{E}[(G(t,X) - G(s,X) \vee G(r,X))^-] \le \epsilon$. I believe this result is originally due to J.A. Yan, in the hard-to-find paper "On the commutability of essential infimum and conditional expectation operators".

Note: $\sup_{t \in [0,1]}G(t,X)$ need not be measurable, which is why the essential supremum is used in the aforementioned theorem.

A more transparent condition can be derived from the above if we add a continuity assumption: Assume $G(\cdot,x)$ is continuous for each $x$, $G(t,\cdot)$ is measurable for each $t$, and $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$. Then $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)]$ if and only if there exists $T \in [0,1]$ such that $G(T,X) = \sup_{t \in [0,1]}G(t,X)$ almost surely.

PROOF: The continuity assumption guarantees that $\sup_{t \in [0,1]}G(t,X)$ is indeed measurable (e.g. by Theorem 18.19 of Aliprantis & Border), and thus $$ \text{ess}\sup_{t \in [0,1]}G(t,X) = \sup_{t \in [0,1]}G(t,X). $$ The aforementioned theorem and a simple argument using compactness of $[0,1]$ and Fatou's lemma shows that (under our assumptions) $$ \sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)] $$ if and only if for all $$ \mbox{$r,s \in [0,1]$ there exists $t \in [0,1]$ such that $G(t,X) \ge G(r,X) \vee G(s,X)$ a.s.} $$ Since the "if" part is trivial, we now prove the "only if". Consider the set $S := \{G(t,X) : t \in [0,1]\}$ with the partial order given by almost sure inequality. Compactness of $[0,1]$ and continuity of $G(\cdot,x)$ for all $x$ yield the existence of an upper bound in $S$ for any chain of $S$, and thus by Zorn's lemma $S$ contains a maximal element. That is, there exists $T \in [0,1]$ such that there is no $s \in [0,1]$ for which $$ \mbox{$G(s,X) \ge G(T,X)$ a.s. and $P(G(s,X) > G(T,X)) > 0$.} $$ For any $t \in [0,1]$ there exists $r \in [0,1]$ such that $$ G(r,X) \ge G(T,X) \vee G(t,X) \ge G(T,X) \mbox{a.s.} \,, $$ which implies $G(r,X) = G(T,X)$ a.s. and thus $G(T,X) \ge G(t,X)$ a.s.. Hence $G(T,X) \ge G(t,X)$ a.s. for any $t \in [0,1]$. $\blacksquare$

This will require very strong conditions on $G$. The most general result I know of is an "almost upward-filtering" condition, fairly well known in stochastic optimal control theory: Assume $G(t,\cdot)$ is measurable for each $t \in [0,1]$, and $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$. Then $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\text{ess}\sup_{t \in [0,1]}G(t,X)]$ if and only if for all $\epsilon > 0$ and $r,s \in [0,1]$ there exists $t \in [0,1]$ such that $\mathbb{E}[(G(t,X) - G(s,X) \vee G(r,X))^-] \le \epsilon$. I believe this result is originally due to J.A. Yan, in the hard-to-find paper "On the commutability of essential infimum and conditional expectation operators". Note that $\sup_{t \in [0,1]}G(t,X)$ need not be measurable, which is why the essential supremum is used in the aforementioned theorem.

A more transparent condition can be derived from the above if we add a continuity assumption: Assume $G(\cdot,x)$ is continuous for each $x$, $G(t,\cdot)$ is measurable for each $t$, and $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$. Then $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)]$ if and only if there exists $T \in [0,1]$ such that $G(T,X) = \sup_{t \in [0,T]}G(t,X)$ almost surely.

PROOF: The continuity assumption guarantees that $\sup_{t \in [0,1]}G(t,X)$ is indeed measurable (e.g. by Theorem 18.19 of Aliprantis & Border), and thus $\text{ess}\sup_{t \in [0,1]}G(t,X) = \sup_{t \in [0,1]}G(t,X)$. The aforementioned theorem and a simple argument using compactness of $[0,1]$ and the dominated convergence theorem shows that (under our assumptions) $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)]$ if and only if for all $r,s \in [0,1]$ there exists $t \in [0,1]$ such that $G(t,X) \ge G(r,X) \vee G(s,X)$ a.s.. Since the "if" part is trivial, we now prove the "only if". Consider the set $S := \{G(t,X) : t \in [0,1]\}$ with the partial order given by almost sure inequality. Compactness of $[0,1]$ and continuity of $G(\cdot,x)$ for all $x$ yield the existence of an upper bound in $S$ for any chain of $S$, and thus by Zorn's lemma $S$ contains a maximal element. That is, there exists $T \in [0,1]$ such that there is no $s \in [0,1]$ for which $G(s,X) \ge G(T,X)$ a.s. and $P(G(s,X) > G(T,X)) > 0$. For any $t \in [0,1]$ there exists $r \in [0,1]$ such that $G(r,X) \ge G(T,X) \vee G(t,X) \ge G(T,X)$ a.s., which implies $G(r,X) = G(T,X)$ a.s. and thus $G(T,X) \ge G(t,X)$ a.s.. Hence $G(T,X) \ge G(t,X)$ a.s. for any $t \in [0,1]$.

This will require very strong conditions on $G$. The most general result I know of is an "almost upward-filtering" condition, fairly well known in stochastic optimal control theory: Assume $G(t,\cdot)$ is measurable for each $t \in [0,1]$, and $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$. Then $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\text{ess}\sup_{t \in [0,1]}G(t,X)]$ if and only if for all $\epsilon > 0$ and $r,s \in [0,1]$ there exists $t \in [0,1]$ such that $\mathbb{E}[(G(t,X) - G(s,X) \vee G(r,X))^-] \le \epsilon$. I believe this result is originally due to J.A. Yan, in the hard-to-find paper "On the commutability of essential infimum and conditional expectation operators". Note that $\sup_{t \in [0,1]}G(t,X)$ need not be measurable, which is why the essential supremum is used in the aforementioned theorem.

A more transparent condition can be derived from the above if we add a continuity assumption: Assume $G(\cdot,x)$ is continuous for each $x$, $G(t,\cdot)$ is measurable for each $t$, and $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$. Then $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)]$ if and only if there exists $T \in [0,1]$ such that $G(T,X) = \sup_{t \in [0,1]}G(t,X)$ almost surely.

PROOF: The continuity assumption guarantees that $\sup_{t \in [0,1]}G(t,X)$ is indeed measurable (e.g. by Theorem 18.19 of Aliprantis & Border), and thus $\text{ess}\sup_{t \in [0,1]}G(t,X) = \sup_{t \in [0,1]}G(t,X)$. The aforementioned theorem and a simple argument using compactness of $[0,1]$ and Fatou's lemma shows that (under our assumptions) $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)]$ if and only if for all $r,s \in [0,1]$ there exists $t \in [0,1]$ such that $G(t,X) \ge G(r,X) \vee G(s,X)$ a.s.. Since the "if" part is trivial, we now prove the "only if". Consider the set $S := \{G(t,X) : t \in [0,1]\}$ with the partial order given by almost sure inequality. Compactness of $[0,1]$ and continuity of $G(\cdot,x)$ for all $x$ yield the existence of an upper bound in $S$ for any chain of $S$, and thus by Zorn's lemma $S$ contains a maximal element. That is, there exists $T \in [0,1]$ such that there is no $s \in [0,1]$ for which $G(s,X) \ge G(T,X)$ a.s. and $P(G(s,X) > G(T,X)) > 0$. For any $t \in [0,1]$ there exists $r \in [0,1]$ such that $G(r,X) \ge G(T,X) \vee G(t,X) \ge G(T,X)$ a.s., which implies $G(r,X) = G(T,X)$ a.s. and thus $G(T,X) \ge G(t,X)$ a.s.. Hence $G(T,X) \ge G(t,X)$ a.s. for any $t \in [0,1]$.