Convergence almost surely implies convergence in probability

Question

Let $X_n\rightarrow^{as}f$ and $N=:\{\omega \in\Omega:X_n(\omega)\nrightarrow X_n(\omega)\}$ so that $P(N)=0$.

I understand why for $\epsilon>0$ and $n\in \mathbb{N}$ we get $$ A_n:=\{\omega \in\Omega:\sup_{m\geq n}\mid X_n(\omega)- X(\omega)\mid\geq \epsilon \} $$

and $A_n \downarrow$ and $\omega \in A_n \forall n\in \mathbb{N}\Rightarrow\forall n\in \mathbb{N}, \exists m\geq n: \mid X_n(\omega)- X(\omega)\mid\geq \epsilon /2$

But why is: $$ A_n \downarrow \bigcap_{m=1}^\infty A_m:= N_\epsilon$$ and also why is $$N_\epsilon \subset \{ \limsup_{n \rightarrow \infty}\mid X_n(\omega)- X(\omega)\mid\geq \epsilon /2\} \subset N $$

Answer 1

For convenience let us work with $Y_{n}:=X_{n}-X$ and for completness let us use the notation $A_{n}^{\left(\epsilon\right)}:=\left\{ \sup_{m\geq n}\left|Y_{m}\right|\geq\epsilon\right\} $ where $\epsilon>0$ is fixed.

Define $N_{\epsilon}:=\bigcap_{n=1}^{\infty}A_{n}^{\left(\epsilon\right)}$.

Then on base of the fact that the sequence of sets $\left(A_{n}^{\left(\epsilon\right)}\right)_{n\in\mathbb{N}}$ is decreasing we are allowed to conclude that $$P\left(A_{n}^{\left(\epsilon\right)}\right)\downarrow P\left(N_{\epsilon}\right)$$

Further we have: $$\left\{ \left|Y_{n}\right|\geq\epsilon\right\} \subseteq A_{n}^{\left(\epsilon\right)}$$ and consequently: $$P\left(\left|Y_{n}\right|\geq\epsilon\right)\leq P\left(A_{n}^{\left(\epsilon\right)}\right)$$

This together tells us that for proving that $\lim_{n\to\infty}P\left(\left|Y_{n}\right|\geq\epsilon\right)=0$ it is enough to prove that $P\left(N_{\epsilon}\right)=0$.

Now observe that: $$\omega\in N_{\epsilon}\iff\forall n\;\omega\in A_{n}^{\left(\epsilon\right)}\iff\forall n\;\sup_{m\geq n}\left|Y_{m}\left(\omega\right)\right|\geq\epsilon\implies$$$$\forall n\exists m\left[m\geq n\wedge\left|Y_{m}\left(\omega\right)\right|\geq\frac{1}{2}\epsilon\right]\implies\omega\in N$$

Proved is now that for every $\epsilon>0$ we have $N_{\epsilon}\subseteq N$ and consequently $P\left(N_{\epsilon}\right)\leq P\left(N\right)$.

So if $P\left(N\right)=0$ then we are allowed to conclude that $P\left(N_{\epsilon}\right)=0$.

This works for every $\epsilon>0$.

Answer 2

$\sup_{m \geq n+1} |X_n(\omega)-X(\omega)| \leq \sup_{m \geq n} |X_n(\omega)-X(\omega)|$ so $A_{n+1} \subset A_n$. Hence $A_n$ decreases to $\cap_n A_n$ which we call $N_{\epsilon}$. $N_{\epsilon} \subset \{\lim \sup |X_n(\omega)-X(\omega)|\geq \epsilon /2\}$ follows from what you have already observed and the fact that $\{\lim \sup |X_n(\omega)-X(\omega)| \}\subset N$ is obvious.