A martingale with bounded increments either converges or diverges to both infinities a.s.

Question

I am reading page 236 "Probability : theory and examples" by R. Durrett. Theorem 31. Let $X_1, X_2,\ldots$ be a martingale with $|X_{n+1}-X_n|\leq M<\infty$. Let

$C=\{\lim X_n \;\;\; \text{exists and finite} \}$

$D=\{\limsup X_n =\infty \;\;\; \text{and}\;\; \liminf X_n =-\infty \}$ then $P(C\cup D)=1$.

Proof

Since $X_n-X_0$ is a martingale , so we can without loss of generality suppose that $X_0=0$. Now let $0<K<\infty$ and let $N=\inf\{n : X_n\leq -K\}$. $X_{n\wedge N}$ is a martingale with $X_{n\wedge N}\geq -K-M$. By applying (2.11) for $X_{n\wedge N}+K+M$, $\lim X_n$ exists on $\{ N=\infty\}$. By letting $K\to \infty$, we see that the limit exists on $\{\liminf X_n>-\infty\}$. $\Box$

My questions are:

1. Why $X_{n\wedge N}\geq -K-M$? isn't it $X_{n\wedge N}\geq -K$ ?

2. Why does the limit exist on $\{\liminf X_n>-\infty\}$ ?

It would be greatly appreciated if some one could help me on this. Thanks for your time.

Answer 1

1. No, $X_{n \wedge N} \geq -K-M$ is correct. Note that, by definition, $X_k \geq -K$ for any $k < N$. Hence, in particular for $k := (N-1) \wedge n$. Consequently, $$X_{n \wedge N} = \underbrace{(X_{n \wedge N}-X_{(N-1) \wedge n})}_{\geq -M} + \underbrace{X_{(N-1) \wedge n}}_{\geq -K} \geq -K-M$$ where we have used that $|X_{k+1}-X_k| \leq M$ for all $k \in \mathbb{N}$.
2. Suppose that $\omega \in \{\liminf X_n>-\infty\}$. Then there exists $K \in \mathbb{R}$ such that $$X_n(\omega)>-K \tag{1}$$ for all $n \in \mathbb{N}$. Indeed: By definition of limit inferior, there exists $m \in \mathbb{N}$ and $r \geq 0$ such that $$X_n(\omega) > -r \quad \text{for all} \, n \geq m.$$ If we set $$K := \max\{|X_1(\omega)|,\ldots,|X_m(\omega)|,r\}+1,$$ then $(1)$ holds. Consequently, $N(\omega)=\infty$ for $N := \inf\{n; X_n \leq -K\}$. Since the limit exists on $\{N=\infty\}$, this finishes the proof.