A question about a UI martingale with 2 stopping times If $L$ and $M$ are stopping times with $L\leq M$ and $Y_{n\wedge M}$ is a uniformly integrable submartingale, then $EY_L\leq EY_M$ and $Y_L\leq E(Y_M|\mathcal{F}_L)$.
Since $Y_{n\wedge M}$ is UI, $Y_{n\wedge M}\rightarrow Y_M$ a.s. and in $L^1$. $Y_{n\wedge M}$ is a submartingale and hence $EY_M\geq EY_{L\wedge M} = EY_L$. I'm not sure how to show $Y_L\leq E(Y_M|\mathcal{F}_L)$. Since, $Y_{n\wedge M}\rightarrow Y_M$ in $L^1$, for $A\in \mathcal{F}_L$, $E(Y_{n\wedge M};A)\rightarrow E(Y_M;A)$ and since $Y_{n\wedge M}$ is monotone in $n$, $E(Y_M;A)\geq E(Y_{L\wedge M};A) = E(Y_L;A)$ but does that mean $E(Y_M|\mathcal{F}_L)\geq EY_L$? It would be true for equality and am not sure about strict inequality? Thanks.
 A: Let's start from $$\mathbb{E}(Y_M 1_A) \geq \mathbb{E}(Y_L 1_A), \qquad A \in F_L \tag{1}$$ which you already proved. Since $$\int_A Y_M \, d\mathbb{P} = \int_A \mathbb{E}(Y_M \mid F_L) \, d\mathbb{P}, \qquad A \in F_L,$$ by the definition of the conditional expectation, it follows from $(1)$ that $$\mathbb{E}(\mathbb{E}(Y_M \mid F_L) 1_A) \geq \mathbb{E}(Y_L 1_A), \qquad A \in F_L.$$ This means that the random variable $Z:=\mathbb{E}(Y_M \mid F_L) - Y_L$ satisfies
$$\mathbb{E}(Z 1_A) \geq 0, \qquad A \in F_L.$$
Since $Z$ is $F_L$-measurable, we have $A:=\{Z \leq 0\} \in F_L$. From $$\int_{\{Z<0\}} Z \, d\mathbb{P} \geq 0,$$ we conclude that $\mathbb{P}(Z<0)=0$. Hence, $\mathbb{E}(Y_M \mid F_L) \geq Y_L$ almost surely.
A: Fix any $A\in\mathcal F_L$, let
$$T(\omega)=\begin{cases}
L(\omega)&\omega\in A,\\
M(\omega)&\omega\in A^c.
\end{cases}$$
It is easy to verify that $T$ is a stopping time such that $T\leq M$ so $EY_T\leq EY_M$. It follows that $E(Y_L1_A)\leq E(Y_M1_A)$ since on $A^c, T=M$.
