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I tried to show that $X_n \xrightarrow{\text{a.s.}} X \Rightarrow X_n \xrightarrow{\text{P}} X$, but my proof is little different from ones I've seen because it works with the limit definition. I feel uneasy about a step in the middle involving selecting the maximum $M$ (Because $M$ might depend on the $\omega$ that we're dealing with, and $\Omega$ could be infinite). I'm also unsure if some of this logic works in general (Like for example putting the limit definition in the probability measure).

Let $X_n \xrightarrow{\text{a.s.}} X$. Then $$\mathbb{P}(\{\omega \in \Omega: X_n(\omega) \rightarrow X(\omega)\}) = 1 \\ \Rightarrow \mathbb{P}(\{\omega \in \Omega: \forall \varepsilon>0, \exists M \in \mathbb{N} \: \text{s.t.} \: \forall m \geq M, \;\; |X_m(\omega) - X(\omega)| < \varepsilon\}) = 1$$

Now let: $$\epsilon > 0, \delta >0. \:\text{Let} \: N \in \mathbb{N}\: (=\max_{\omega \in \Omega}{M_{\epsilon}(\omega)}). \forall n \geq N, \forall \omega \in \Omega \setminus \mathcal{N}, |X_n(\omega) - X(\omega)| < \epsilon \: \text{(Drawing on above)}$$

Where $\mathcal{N}$ denotes the union of $\mathbb{P}$-null sets.

Thus: $$\mathbb{P}(\{\omega \in \Omega : |X_n(\omega)-X(\omega)| > \epsilon \}) = \mathbb{P}(\{\omega \in \mathcal{N}: |X_n(\omega)-X(\omega)| > \epsilon \}) + \mathbb{P}(\{\omega \in \Omega \setminus \mathcal{N}: |X_n(\omega)-X(\omega)| > \epsilon \}) = 0 < \delta$$

And thus we've shown: $$\forall \epsilon > 0, \lim_{n \rightarrow \infty} \mathbb{P}(\{\omega \in \Omega :|X_n(\omega) - X(\omega)| > \epsilon\}) = 0 \Rightarrow X_n \xrightarrow{\text{P}}X$$

If the above doesn't work, is it possible to show this somehow using the limit construction formally? Thanks.

rudinable
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    Not a full answer, but just an initial comment: what exactly do you mean by P-null? If you mean the union of all sets that individually have measure 0, that's a problem - the union of all P-null sets in $\mathbb{R}$ is $\mathbb{R}$! Indeed, just take the union of all singletons. (note this will require taking an uncountable union, but again, over the reals there are an uncountable number of P-null sets) – masiewpao Jan 20 '25 at 14:28
  • Null sets on the sigma algebra for which $\mathbb{P}$ is defined here. Because we're only given convergence almost surely, then the property inside the probability measure only holds for certain for any $\omega$ which is not contained within a null set of the sigma algebra. Am I right to say this? – rudinable Jan 20 '25 at 14:52
  • No, not necessarily. Think about the singletons of $\mathbb{R}$ - these are all null sets within the sigma algebra (I'm assuming the Borel sigma algebra here). But if you were to take the (uncountable) union of them all, you are left with a set of measure 1. – masiewpao Jan 20 '25 at 15:01
  • Ah, you're right. But in that case, how can we formalize that the information that we're given only applies almost everywhere, and not necessarily for all $\omega$? – rudinable Jan 20 '25 at 15:02
  • Also another small comment - you are right about your concern for about selecting the maximum $M$. In the general case, you cannot guarantee a single, finite, bound will exist for all $\epsilon$. – masiewpao Jan 20 '25 at 16:20

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