Related to this question, @AlexR gave a perfect answer, but left a non-trivial statement as an ending to the proof. He said that
$\forall\epsilon>0 \ ,\ \sum\limits_1^\infty P(|M_n-x^*_F|>\epsilon)<\infty\implies M_n\xrightarrow{a.s.} x^*_F$
I have tried to prove it. Have I gone wrong anywhere or assumed something wrong?
$$\forall\epsilon>0 \ ,\ \sum\limits_1^\infty P(|M_n-x^*_F|>\epsilon)<\infty\\\implies \forall\epsilon>0 \ ,\ P(\limsup_{n\to\infty}|M_n-x^*_F|>\epsilon)=0\ \ \cdots\cdots(Borel-Cantelli)\\\implies \forall\epsilon>0 \ ,\ P(|M_n-x^*_F|>\epsilon \text{ for infinitely many } n )=0\\\implies P(M_n\to x^*_F \text{ as } n\to\infty)=1\\\implies M_n\xrightarrow{a.s.} x^*_F$$
EDIT:
If the above is true, then I may assume
If $X_n$ is a sequence of iid random variables and $X$ is another random variable such that $\forall\epsilon>0 \ ,\ \sum\limits_1^\infty P(|X_n-x|>\epsilon)<\infty\implies X_n\xrightarrow{a.s.} X$
But Suppose $Y_n$ is a sequence of iid random variables converging in probability to $Y$ , then $$\lim_{n\to\infty}P(|Y_n-Y|>\epsilon)=0\ \ \forall \epsilon>0\\\implies P(\lim_{n\to\infty}|Y_n-Y|>\epsilon)=0\ \ \forall \epsilon>0\\\implies P(\limsup_{n\to\infty}|Y_n-Y|>\epsilon)=0\ \ \forall \epsilon>0\\\implies Y_n\xrightarrow{a.s.} Y \ \ (contradiction)$$
Where am I going wrong?