Is this last statement a convergence in probability, why does this converse holds? 
Theorem 7.6
I can see why does almost sure convergence implies convergence in probability, and I don't see why does this converse also holds.
Thank you for any help
Thank you all for the posts, here it does not say anything about the nested sequence, so I did not think about it this way. If increasing sequence is provided, is that sufficient prove this solely by the continuity of probability?
 A: $(A_m)$ is an increasing sequence.  So $\lim P(A_m)=P(A)$ where $A =\bigcup_m A_m =\{|X_n-X| <\epsilon \, \, \text {for all }\, n \,\,\text {sufficiently large}\}$. We can let $\epsilon \to 0$ through a sequence and conclude that $1=P(A)=P(X_n \to X)$.
A: This just provides more details to geetha290krm's answer.
$(\Leftarrow)$. Note that $A_{\varepsilon,m}=\cap_{n\geq m}\{|X_n-X|<\varepsilon\}$ and $A_{\varepsilon,m}\subseteq A_{\varepsilon ,m+1}$. Then define $A_\varepsilon:=\cup_{m \in \mathbb{N}}A_{\varepsilon,m}$ (it is a monotone class; this also implies $P(A_\varepsilon)=\lim_mP(A_{\varepsilon,m})$ by measure continuity). We have
$$A_\varepsilon=\bigcup_{m \in \mathbb{N}}\bigcap_{n\geq m}\{|X_n-X|<\varepsilon\}$$
Now note that this means that if $\omega \in A_\varepsilon$, then there exists $m$ s.t. $\forall n\geq m$, $|X_n(\omega)-X(\omega)|<\varepsilon$. We suppose $P(A_\varepsilon)=1$ for all $\varepsilon >0$. Now consider
$$A:=\bigcap_{k \in \mathbb{N}}A_{1/k}$$
Then $A=\{\lim_n X_n=X\}$. So by De Morgan's laws and measure monotonicity, $P(A^c)\leq \sum_kP(A_{1/k}^c)=0$. So $P(A)=1$ and we conclude that $X_n\to X$ $P$-a.s.
$(\Rightarrow)$. For any $\varepsilon>0$, we have $A\subseteq A_\varepsilon$, so $1=P(A)\leq P(A_\varepsilon)=1$ necessarily.
