Show that :$ \int_{A} X_n d\mathbb{P} \leq \int_{A} X_m d\mathbb{P} $

Question

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, $\left(X_n\right)_{n\ge 1} $ is a sequence of real random variables in $L^{1}(\Omega,\mathcal{F},\mathbb{P})$ and $\left(\mathcal{F}_n\right)_{n\ge 1} $ is an increasing sequence of sub- $\sigma$ -algebra of $\mathcal{F}$ . such as :

$(a)$. $\forall n\ge 1$, $X_n $ is $\mathcal{F}_n$-measurable function.

$(b)$. $\forall n\ge 1$, $\mathbb{E}[X_{n+1}\mid \mathcal{F}_n]\ge X_n ~~a.s.$

Then : $\forall n\ge 1, \forall m\ge n $ : $$ \mathbb{E}[X_{m}\mid \mathcal{F}_n]\ge X_n ~~a.s. $$

Show that : $\forall n\ge 1, \forall m\ge n, \forall A\in \mathcal{F}_n $: $$ \int_{A} X_n d\mathbb{P} \leq \int_{A} X_m d\mathbb{P} $$

Answer 1

You have $\int_{A} X_m\, d\mathbf{P} = \mathbf{E}\left[ \mathbf{1}_A X_m\right] = \mathbf{E}\left[ \mathbf{1}_A \mathbf{E}\left[ X_m \mid \mathscr{F}_n\right]\right] \geq \mathbf{E}\left[ \mathbf{1}_A X_n\right] = \int_{A} X_n\, d\mathbf{P}$ where

• the first and last equalities result from the definition of the integral;
• the second equality results from the definition of conditional expectation w.r.t. $\mathscr{F}_n$ as $\mathbf{1}_A$ is $\mathscr{F}_n$-measurable as $A \in \mathscr{F}_n$;
• the inequality results from the hypothesis and from the fact that if $X \geq Y$ $\mathbf{P}$-a.s. then $\mathbf{E}\left[ X \right] \geq \mathbf{E}\left[ Y \right]$.