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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?

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what is your definition of almost sure convergence – Koushik Dec 25 '12 at 9:07
@K.Ghosh, it is the usual definition. $P\,(\lim_{n \to \infty} X_n = X) = 1$. ( – JDL Dec 25 '12 at 9:10
This is not true. Take $X_{n}=\frac{1}{n}$ for all $n\in\mathbb{N}$. Then $X_{n}\to 0=:X$‚ but $P(X_{n}=X)=P(\frac{1}{n}=0)=0$ for all $n\in\mathbb{N}$. Hence $\lim_{n\to\infty}P(X_{n}=X)=0$. – T. Eskin Dec 25 '12 at 9:11
up vote 2 down vote accepted

The claim is not true. Consider the probability space $([0,1],\mathcal{B}[0,1],\lambda|_{[0,1]})$ (where $\mathcal{B}$ denotes the borel-$\sigma$-algebra, $\lambda|_{[0,1]}$ the lebesgue measure on $[0,1]$) and

$$X_n(\omega) := \frac{1}{n} \qquad (\omega \in [0,1])$$

Then $X_n \to 0=:X$ almost surely, but $$\mathbb{P}[X_n=X] = \mathbb{P}[X_n=0]=0$$

thus clearly $\mathbb{P}[X_n = X] \not \to 1$ as $n \to \infty$.

share|cite|improve this answer
Yes, I think I have misundersood the problem. Anyway, it seems that my derivation with the dominated convergence theorem is also not true. Your example is a counter example for the derivation. Could you point out which line of the equation is wrong? – JDL Dec 25 '12 at 9:32
Yes. $\chi_{\{\omega; \lim_n X_n(\omega)=X_n(\omega)\}} \not= \lim \chi_{\{\omega; X_n(\omega)=X(\omega)\}}$... The example above shows that even if $X_n(\omega) \to X(\omega)$ there does not necessarily exist $N \in \mathbb{N}$ such that $X_N(\omega)=X(\omega)$. – saz Dec 25 '12 at 9:37

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