a problem about martingale, to prove $X_n \le E(X_\infty|B_n)$ Suppose $\{(X_n, B_n), n\ge0\}$ is an $L_1$-bounded martingale. If there exists an integrable random variable Y such that $X_n \le E(Y|B_n)$ then $X_n \le E(X_\infty|B_n)$ for all $n \ge0$ where $X_\infty=\lim_{n\to \infty}X_n$ almost surely.
I know $X_n$ converges to $X_\infty$ a.s.,but I really cannot link the condition $X_n \le E(Y|B_n)$ to X. Any hints? Martingale is really too hard for me..
 A: Define a cut-off function
$$\chi_k(x) := \begin{cases} x & x \geq -k \\ -k & x<-k \end{cases}$$
for $k \in \mathbb{N}$. Obviously, $\chi_k$ is convex and therefore, by Jensen's inequality, $(\chi_k(X_n))_{n \in \mathbb{N}}$ is a submartingale. Suppose we knew that $(\chi_k(X_n))_{n \in \mathbb{N}}$ is uniformly integrable. Then $X_n \to X_{\infty}$ a.s. implies $\chi_k(X_n) \to \chi_k(X_{\infty})$ as $n \to \infty$ almost surely and in $L^1$. Using the submartingale property, we find
$$\int_F \chi_k(X_n) \, d\mathbb{P} \leq \int_F \chi_k(X_{\infty}) \, d\mathbb{P}$$
for any $F \in \mathcal{B}_n$. Since $X_n$, $X_{\infty} \in L^1$, the dominated convergence theorem yields
$$\int_F X_n \, d\mathbb{P} \leq \int_F X_{\infty} \, d\mathbb{P}$$
and this finishes the proof.


Lemma: $(\chi_k(X_n))_{n \in \mathbb{N}}$ is uniformly integrable.


Remark: Basically, the proof makes use of the following two statements:


*

*For any $Y \in L^1$, the family of random variables $\{\mathbb{E}(Y \mid \mathcal{F}); \mathcal{F} \subseteq \mathcal{A}$ is a $\sigma$-algebra$\}$ is uniformly integrable where $\mathcal{A}$ denotes the $\sigma$-algebra of the underlying probability space.

*Let $(X_n)_n$, $(Y_n)_n$ such that $|X_n| \leq Y_n$ and $(Y_n)_n$ is uniformly integrable. Then $(X_n)_n$ is uniformly integrable.


Proof: First of all, we note that it suffices to show that the positive and negative part of $(\chi_k(X_n))$ are uniformly integrable. Since $\chi_k(X_n)$ is bounded below, the uniformly integrability of the negative part is obvious. It remains to show that the positive part, i.e. $$\max\{\chi_k(X_n),0\} = X_n^+,$$ is uniformly integrable.
By assumption, we have $X_n^+ \leq \mathbb{E}(Y \mid \mathcal{B}_n)=:Y_n$, hence
$$\int_{|X_n^+|>R} X_n^+ \, d\mathbb{P} \leq \int_{|Y_n| >R} |Y_n| \, d\mathbb{P}. $$
Consequently, it suffices to show that $(Y_n)_n$ is uniformly integrable. From $$|Y_n| = |\mathbb{E}(Y \mid \mathcal{B}_n)| \leq \mathbb{E}(|Y| \mid \mathcal{B}_n)$$ we conclude
$$\begin{align*} \int_{|Y_n|>R} |Y_n| \, d\mathbb{P} &\leq \int_{|Y_n|>R} \mathbb{E}(|Y| \mid \mathcal{B}_n) \, d\mathbb{P} = \int_{|Y_n|>R} |Y| \, d\mathbb{P} \\ &\leq \int_{|Y_n|>R, |Y| \leq K} |Y| \, d\mathbb{P} + \int_{|Y_n|>R, |Y|>K} |Y| \, d\mathbb{P} \\ &\leq K \cdot \frac{\mathbb{E}|Y|}{R} + \int_{|Y|>K} |Y| \, d\mathbb{P} \end{align*}$$
where we applied Markov's inequality in the last step. Letting first $R \to \infty$ and then $K \to \infty$ proves the claim.
