Convergence in probability and almost surely Let $X_n$ be a sequence of independent random variable which converges in probability to $X$. Prove $X$ is a constant.
Can someone give me a hint how I should go about proving this? I tried proving this by contradiction by saying $X$ taking 2 different values, but this can still still happen because $X$ is only almost surely constant.
 A: Hint 1: Passing to a subsequence, we can assume $X_n \to X$ almost surely.
Hint 2: Do you know the Kolmogorov zero-one law?  $X$ is a tail random variable.
A: Maybe not rigorous, but the idea is correct.
$$ X_n\xrightarrow{a.s}X\Leftrightarrow P(\lim_{n\rightarrow\infty}\lvert |X_n-X\rvert |>\epsilon)=0$$
$$\lvert|X_n-X\rvert|+\lvert|X_{n+1}-X\rvert|\geq\lvert|X_n-X-X_{n+1}+X\rvert|=\lvert|X_n-X_{n+1}\rvert|$$$$\Rightarrow\lim_{n\rightarrow\infty}\lvert|X_n-X\rvert|+\lim_{n\rightarrow\infty}\lvert|X_{n+1}-X\rvert|\geq\lim_{n\rightarrow\infty}\lvert|X_n-X_{n+1}\rvert|$$$$\Rightarrow 0=P(\lim_{n\rightarrow\infty}\lvert|X_n-X\rvert|>\epsilon/2)+P(\lim_{n\rightarrow\infty}\lvert|X_{n+1}-X\rvert|>\epsilon/2)\geq P(\lim_{n\rightarrow\infty}\lvert|X_n-X\rvert|>\epsilon/2\lor\lim_{n\rightarrow\infty}\lvert|X_{n+1}-X\rvert|>\epsilon/2)$$$$=P(\lim_{n\rightarrow\infty}\lvert|X_n-X\rvert|+\lim_{n\rightarrow\infty}\lvert|X_{n+1}-X\rvert|>\epsilon)$$$$\geq P(\lim_{n\rightarrow\infty}\lvert|X_n-X_{n+1}\rvert|>\epsilon)\geq0$$
Since $X_n$s are independent, $X_n$ converges to a constant almost surely.
