We have random variables $X_n\rightarrow X$ almost surely. Now fix a positive real number $M$, define $\bar{Y}=Y \mathbb{I}_{\{Y\leq M\}}$. Then if $P(X=M)=0$, we have $\bar{X_n}\rightarrow \bar{X}$ almost surely.
My question is why we need $P(X=M)=0$ this condition? And would someone please give me a counterexample that this convergence fails without this condition?
Thanks
Let $X_n=M+1/n$ and $X=M$; then $\overline{X_n}=0$ and $\overline{X}=M$ while $X_n\to X$ almost surely.
The reason is that if $P(X=M)=0$, then for almost every $\omega$, we either have $X(\omega)<M$ or $X(\omega)>M$ hence for $n$ large enough, we are sure that $X_n(\omega)<M$ or $X_n(\omega)>M$ and the value of the indicator will not change.