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Let $\{X_n\}_{n=1}^\infty$ be a sequence of iid random variables over a probability space $(\Omega,\mathcal{F},\mathrm P)$. Prove that $$\mathrm P\left(\lim_{n\to\infty} X_n \text{ exists}\right)=1$$ iff the distribution $\mu$ of $X_1$ is degenerate, i.e., $\mu(A)\in\{0,1\}$ for all $A\in\mathcal{B}(\mathbb R)$, the usual Borel-measurable set.

What I have done:

I'm acquainted with the Kolmogorov's Zero-One law, then if $A\in\mathcal{T}$, then $P(A)\in\{0,1\}$. Where $\mathcal{T}$ is the $\sigma-$tail generated by the iid random variable sequence $X_n$. In this case $\sigma(X_1)=\mathcal{T}$.

I proved that $\{\lim X_n\quad\mathrm{exists}\}\in\sigma(X_1)$, therefore $\mathrm{P}(\{\lim X_n\quad\mathrm{exists}\})\in\{0,1\}$.

My intuition is if I prove that $\mathrm{P}(\{\lim X_n=c\})=1$ (for some $c\in\overline{\mathbb{R}}$), then it follows directly that $\mu$ is degenerate. I found in Shiryaev's Probability (second edition) that if $Y$ is $\mathcal{T}$-measurable, then $\mathrm{P}(\{Y=k\})$ is precisely 1 (for some $k\in\overline{\mathbb{R}}$), but I don't know how to prove the previous claim, but I know that $\lim X_n$ is $\mathcal{T}$-measurable.

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marked as duplicate by Community Sep 29 '15 at 13:07

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  • $\begingroup$ Thanks with the edition @Math1000 is a neat question now :) . $\endgroup$ – DonQuixote Sep 27 '15 at 3:23