# Convergence of random variables implying almost sure convergence (where is the flaw?)

This would probably look like a dumb question (akin to prove $2=1$ type). But I'd still like to know where the flaw lies. The question is regarding convergence in probability and almost sure convergence. We know the following: Suppose events $A_n \to A$, then $P(A_n) \to P(A)$ i.e $\lim P(A_n) = P(\lim A_n) = P(A)$.

Can't the same principle be used to conclude equivalence of both type of convergence?

$P(\lim X_n = X) = \lim P(X_n = X)$, implying equivalence? Thanks!

Best,

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For some basic information about writing math at this site see e.g. here, here, here and here. – Julian Kuelshammer Oct 30 '12 at 7:47
Thanks for the edits!... sorry for being (extremely) sloppy at first! – MMM Oct 30 '12 at 7:57

## 1 Answer

If you write the parentheses in your second line instead of being sloppy, you see what happens:

In your first example, you use that if $A_n \to A$, then $P(A_n) \to P(A)$. If you want to use that for $\{X_n = X\}$ as you do, you must have $\{X_n = X\} \to \{X=X\} = \Omega$ or something like that. See also that in your first example, the right hand side $P(A)$ doesn't contain a limit, but the one in the second does. Please be more exact.

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Sorry for the typo in the first example...its corrected now...thanks for the answer! – MMM Oct 30 '12 at 7:59