prove that almost surely implies convergence in measure

Asked Dec 31 '19 at 02:45

Active Dec 31 '19 at 15:57

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Prove that:

if convergence almost surly(that is $P(\lim_nX_n = X) = 1$) to a finite limit implies convergence in measure(that is $P(|X_n-X|>\epsilon) \to 0$).

My attempt:

For convergence almost surely we can find a set $E$ with it's complement measure zero.On set $E$ we have $X_n \to X$ pointwise.

Hence by Egroff theorem we can find a subset of $U\subset E$ that $X_n$ convergence uniformly on $U$(with $E-U$ measure be arbitrary small $\eta$).

Hence For for any $\epsilon$ chosen by $|X_n-X|>\epsilon$.We have for any $\eta>0$ a $N(\epsilon) $ such that when $n>N$ ,we have$|X_n-X|>\epsilon$ can not happen on U.with it's complement measure arbitary small.

hence we have $P(\{|X_n-X|>\epsilon\})<P(X-E)+P(E-U) = \eta) $ which is arbitray small.hence convergence in measure

Is my proof correct?

A relavant converse case see here :converse case

edited Dec 31 '19 at 03:09

asked Dec 31 '19 at 02:45

yi li

4,786

There is a standard proof which does not use Egoroff's Theorem and I am not sure if this proof would be considered as a reasonable one. – Kavi Rama Murthy Dec 31 '19 at 05:34
Can you provide a reference about the standard proof,since I don't have it at my hand. – yi li Dec 31 '19 at 06:02
Every book on measure theoretic Probability Theory has a proof. Examples: books of Chung, Breiman etc. – Kavi Rama Murthy Dec 31 '19 at 06:03
Maybe it's correct since convergence almost uniformly implies convergence in measure – yi li Dec 31 '19 at 08:31

prove that almost surely implies convergence in measure

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