This is a homework problem. The problem is proving that $\lim_{n\to\infty}P\,(X_n=X) = 1$ when the sequence of random variables $X_n$ converges to $X$ almost surely.
I think the problem is essentially about the order of the probability sign and limit sign. I know that the order cannot be interchanged in general. However, if $X_n$ converges a.s. and each $X_n$ is dominated by an integrable random variable $Y$, in the light of the dominated convergence theorem, $ \begin{eqnarray} 1 = P\,(\lim_{x\to\infty}X_n=X) &=& \int_{\{\omega\,:\,\lim_{n\to\infty}X_n(\omega)=X(\omega)\}} d\mu \\ &=& \int\chi_{\{\omega\,:\,\lim_{n\to\infty}X_n(\omega)=X(\omega)\}} d\mu \\ &=& \int\lim_{n\to\infty}\chi_{\{\omega\,:\,X_n(\omega)=X(\omega)\}} d\mu \\ &=& \lim_{n\to\infty}\int\chi_{\{\omega\,:\,X_n(\omega)=X(\omega)\}} d\mu \\ &=& \lim_{n\to\infty}P\,(X_n=X) \end{eqnarray} $
I this derivation right? But, even if it is right, I have no idea how to prove it without the dominance condition. How can I prove it generally? Is it possible actually?