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Ok, so Im having a hard time in trying to find the essential difference between convergence in probability and almost sure convergence. So I've got that

$$X_n \overset{P}\to X \iff \forall \epsilon >0, \lim_{n\to \infty}\Bbb{P}(|X_n - X|\geq \epsilon)=0 \iff \forall \epsilon >0, \exists n_0 \text{ such that } \Bbb{P}(|X_n - X|\geq \epsilon)<\epsilon \text{ for all } n\geq n_0$$

While

$$X_n \overset{a.s}\to X \iff \Bbb{P}(\{\omega:X_n(\omega)\to X(\omega)\})=1 \iff \forall \epsilon >0, \lim_{m\to \infty}\Bbb{P}(\bigcup_{n=m}^{\infty}\{|X_n - X|\geq \epsilon\})=0$$

So Im not being able to see the $essential$ difference between this two types of convergence, were can I see that Almost Sure convergence is stronger? Can anyone clarify this to me?

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I find it instructive to consider the case $X_n =1_{M_n}$ (i.e. each $X_n$ is an indicator function).

You should then verify the two claims below:

$X_n$ converges in measure to 0 if and only if $\Bbb{P}(M_n)\to 0$.

$X_n$ converges almost surely to 0 if and only if $\Bbb{P}(\bigcup_{k=n}^\infty M_k)\to 0$ as $n\to\infty$.

Finally, you should know the abstract results that (at least for probability measures), almost sure convergence implies convergence in measure and convergence in measure implies almost everywhere convergence for a subsequence.

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