Martingale $M_n$ - Show that $\mathbb{P}(\{ \tau = \infty\}\backslash \{\lim\limits_{n \rightarrow \infty}{M_n ~\mathrm{exists}}\}) = 0$

Question

Given $(M_n)_{n \in \mathbb{N_0}}$ a martingale such that $\exists K < \infty$ with $|M_n - M_{n-1}| < K ~\forall n \in \mathbb{N}$, and the events $A=\{ \lim\limits_{n \rightarrow \infty}{M_n ~\mathrm{exists}}\}, ~B=\{ \limsup\limits_{n \rightarrow \infty}{M_n}=\infty ~\mathrm{and}~\liminf\limits_{n \rightarrow \infty}{M_n} = -\infty\}$.

We have $L \in (0,\infty)$ and a stopping time $\tau = \inf\{n: M_n \leq -L \}$ and I want to show that $\mathbb{P}(\{ \tau=\infty \} \backslash C)=0 $

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My idea was: $$ \{ \tau = \infty\} = \{ M_1 > -L\} \cap \{M_2 > -L\} \cap \dots $$ which means, for all $n \in \mathbb{N}: ~ M_n > -L$.

Now consider set $A$. $$A = \{ \lim\limits_{n \rightarrow \infty}{M_n ~\mathrm{exists}}\}= \{- \infty < \liminf\limits_{n \rightarrow \infty}M_n = \limsup\limits_{n \rightarrow \infty}M_n < \infty\}$$ Taking the complement, we have: $$\Omega\backslash C = \{ \liminf\limits_{n \rightarrow \infty}M_n \neq \limsup\limits_{n \rightarrow \infty}M_n\} \cup \{ \liminf\limits_{n \rightarrow \infty}M_n = \limsup\limits_{n \rightarrow \infty}M_n = \infty\}\cup \{ \liminf\limits_{n \rightarrow \infty}M_n = \limsup\limits_{n \rightarrow \infty}M_n = -\infty\}$$

And then we have: $\{ \tau = \infty \} \backslash C = \{ \tau = \infty \}\cap (\Omega \backslash C)$. I am not sure what this set contains, I think it has to be: $\{ \liminf\limits_{n \rightarrow \infty}M_n \neq \limsup\limits_{n \rightarrow \infty}M_n\} \cup \{ \liminf\limits_{n \rightarrow \infty}M_n = \limsup\limits_{n \rightarrow \infty}M_n = \infty\}$ and now I don't know how to show that this set is a $\mathbb{P}$-null set.

Answer 1

Given that $\{ \tau = \infty \}$ you come to the conclusion that $M_n > -L$ for all $n$. Now since $M_n$ is a martingale, $-M_n$ is also martingale. Use that $-M_n < L$, take expectation on both sides and use the martingales convergence theorem to conclude that $-M_n$ converges. Then $M_n$ converges thus $\{ \tau = \infty \} \Rightarrow $ limit exists. Thus the probability must be zero.

Answer 2

1. The stopped martingale $Y_n:=M_{\tau\wedge n}$ satisfies $Y_n\ge -L-K$ (consider the cases $n<\tau$ and $n\ge\tau$ separately). Therefore $\Bbb E[Y_n^-]\le L+K$ for all $n$.

2. $$\Bbb E[|Y_n|]=\Bbb E[Y_n]+2\Bbb E[Y^-_n]=\Bbb E[Y_0]+2\Bbb E[Y^-_n]\le\Bbb E[Y_0]+2(L+K) $$ for all $n$. Thus $(Y_n)$ is $L^1$-bounded, hence convergent.

3. Finish by asking yourself: How are $(Y_n)$ and $(M_n)$ related on the event $\{\tau=\infty\}$?