Prove that $\{\omega \mid X_n (\omega) \to X (\omega)\}$ is measurable

Question

I'm reading the definition of almost-sure convergence.

Let $X$ be a random variable and $\left\{X_{n}\right\}$ be a sequence of random variables on the probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Then $X_{n} \rightarrow X$ a.s. if $\mathbb{P}\left(\left\{\omega \in \Omega: X_{n}(\omega) \rightarrow X(\omega)\right\}\right)=1$.

Could you verify my proof of $\{\omega \mid X_n (\omega) \to X (\omega)\} \in \mathcal F$ is correct?

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We have $X_n (\omega) \to X (\omega)$ iff $\forall \varepsilon >0, \exists N \in \mathbb N, \forall n \ge N: |X_n (\omega) - X (\omega)| < \varepsilon$ iff $\forall m \in \mathbb N^*, \exists N \in \mathbb N, \forall n \ge N: |X_n (\omega) - X (\omega)| < 1/m$. Hence $$\{\omega \mid X_n (\omega) \to X (\omega)\} = \bigcap_{m \in \mathbb N^*} \bigcup_{N \in \mathbb N} \bigcap_{n \ge N} (X_n-X)^{-1} \left ( \left ( -\frac{1}{m},\frac{1}{m} \right) \right).$$

We have $X, X_n$ are measurable, so is $X_n-X$ and thus $(X_n-X)^{-1} \left ( \left ( -\frac{1}{m},\frac{1}{m} \right) \right) \in \mathcal F$. The result then follows.

Answer 1

The solution is perfectly fine (just as a suggestion for the next time questions as this, maybe add the tag “proof verification”).