Let $X$ be a square-integrable martingale with $\left| X_n - X_{n-1}\right| \leq L$ a.s. for some $L > 0$. Let $\langle X \rangle$ be its square variation process $$ \langle X \rangle_n = \sum_{i=1}^n \mathbb E\left[\left(X_i - X_{i-1}\right)^2 \, \big| \, \mathcal F_{i-1}\right] \leq nL^2 $$ so that $X^2 - \langle X \rangle$ is a martingale. I want to show that if $X$ converges a.s., or equivalently (mod $\mathbb P$) if $\displaystyle \limsup_{n \to \infty} X_n <\infty$ a.s., then $\displaystyle \sup_{n \in \mathbb N} \langle X \rangle_n < \infty$ a.s.
My strategy: I want to use the following theorem:
Let $X$ be a square-integrable martingale with square variation process $\langle X \rangle$. Then the following four statements are equivalent:
- $\displaystyle \sup_{n \in \mathbb N} \mathbb E\left[X^2_n\right] < \infty$ (i.e. $X$ is $L^2$-bounded)
- $\displaystyle \lim_{n \to \infty} \mathbb E \left[ \langle X \rangle_n\right] < \infty$
- $X$ converges in $L^2$
So if I can show $X$ either is convergent in $L^2$, or is bounded in $L^2$, I'll be done. I thought I could show $X$ is Cauchy in $L^2$; since $X$ is a martingale, if $n > m$, $$ \mathbb E\left[\left(X_n - X_m\right)^2 \right] = \mathbb E\left[X^2_n - X^2_m\right] $$ but I'm not sure where to go from here. And since $X$ isn't bounded by an integrable function that I know of (its a.s. limit need not be integrable, a priori), I can't use dominated convergence to show that $X$ is $L^2$-bounded. Any suggestions?
EDIT: This answer isn't exactly an answer to this question specifically, but it gets us to the solution since one can show $\{X_n \textrm{ converges}\} = \{\limsup X_n < \infty\} = \{\liminf X_n > -\infty\} \mod \mathbb P$.
