Show that if $X_n$ converges to $X$, also $|X_n|$ converges to $|X|$ almost surely.

Question

Show that if $X_n$ converges to $X$, also $|X_n|$ converges to $|X|$ almost surely.

I know the fact that this will surely hold in the case of the convergence in probability, but I don't know how to prove it in the case of the convergence almost surely.

Answer 1

Assume that $(X_n)_{n \in \mathbb{N}}$ converges to $X$ almost surely. This means that $$P\left(\omega : \lim_{n\to\infty}X_n(\omega) = X(\omega)\right) = 1.$$ Now, we just have to show that $$\{\omega : \lim_{n\to\infty} X_n(\omega) = X(\omega)\} \subset \{\omega : \lim_{n\to\infty} \vert X_n(\omega)\vert = \vert X(\omega)\vert\}$$ in order to conclude that the set in the right hand side has measure $1$ as well. In fact, if $\omega$ is such that $\lim_{n\to\infty}X_n(\omega) = X(\omega)$, using the triangle inequality $$\vert \vert X(\omega) \vert - \vert X_n(\omega) \vert \vert \leq \vert X(\omega) - X_n(\omega) \vert$$ we conclude that $\lim_{n\to\infty} \vert X_n(\omega) \vert = \vert X(\omega)\vert$ and the proof is done.

Answer 2

Almost sure convergence means that there is a set $\Omega_0$ with $\mathbb{P}(\Omega_0) = 1$ such that restricted to $\Omega_0$, $X_n$ converges pointwise to $X$.

Thus all properties enjoyed by pointwise convergence will transfer over to almost sure convergence. In particular, you have the continuous mapping theorem :

If $X_n \to X$ pointwise and $f : \mathbb{R} \to \mathbb{R}$ is continuous, then $ f \circ X_n \to f \circ X $ pointwise

Can you see how this immediately transfers to almost sure convergence?