a.e. convergence to a constant implies the sup of the sequence is almost surely finite

Question

I'd like to show that if $\{X_n\}$ is a sequence of random variables for which $|X_n|\overset{a.e.}\to c$ for some $c>0$, then for every $\epsilon>0$, there exists $M<\infty$ so that $P(\sup_n |X_n|\le M)\ge 1-\epsilon$.

We know $X_n \overset{a.e.}\to c$ implies that $P(|X_n-c|>1\ i.o.)=0$, so $P(|X_n|\le c+1\ i.o.)=1$. So there exists a set $N\subset\Omega$ so that $P(N)=0$ and for all $\omega\in N^c$, $|X_n(\omega)|>c+1$ for only finitely many indices, i.e. there exists some $N(\omega)\in\mathbb{N}$ so that $n\ge N(\omega)$ implies $|X_n(\omega)|\le c+1$. So $\sup_n|X_n(\omega)|\le\max\{|X_1(\omega),...,X_{N(\omega)}(\omega)|,c+1\}\equiv M(\omega)$.

We want to say that there exists some $M$ independent of $\omega$ so that $P(\sup_n |X_n|\le M)\ge 1-\epsilon$. I'm thinking $\sup_{\omega\in N^c}M(\omega)$ is a good candidate, but I'm not sure how to show that it's finite.

Any pointers would be greatly appreciated! Thanks in advance.

Answer 1

For each fixed $\omega \in \Omega$, the sequence $|X_n(\omega)|$ converges and therefore $$\sup_{n \in \mathbb{N}} |X_n(\omega)|<\infty. \tag{1}$$ If we define $$A_R := \left\{\omega \in \Omega; \sup_{n \in \mathbb{N}} |X_n(\omega)| \leq R \right\}$$ then $A_R \subseteq A_S$ for $R \leq S$ and, by $(1)$, $\bigcup_{R \geq 1} A_R = \Omega$. Using the continuity of the probability measure $\mathbb{P}$, we get

$$\lim_{R \to \infty} \mathbb{P}(A_R)=1.$$

In particular, we can choose $R>0$ sufficiently large such that

$$\mathbb{P}(A_R) = \mathbb{P} \left( \sup_{n \in \mathbb{N}} |X_n| \leq R \right) \geq 1-\epsilon$$

for given $\epsilon>0$.