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On this wikipedia page, "setwise convergence" is holds for a sequence of measures as $\mu_n$ to $\mu$ when $\mu_n(A)\to\mu(A)$ for all measurable sets $A$. The article says that "In a measure theoretical or probabilistic context setwise convergence is often referred to as strong convergence." On the page for convergence of random variables "strong convergence" is defined as the same as almost sure convergence. How are these related? For almost sure convergence I picture a sequence of functions $f_n$ converging to $f$ on some measure space $(X,F,\nu)$, like in real analysis. Does this mean $\nu(f_n\in A)\to \nu(f\in A)$ for all $A$ measurable in the image space iff $f_n\to f$ a.s? Edit: If not, what can it mean?

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No. $ \forall A$ s.t. $A$ is measurable in the image space $v(f_{n} \in A) \to v(f \in A)$ does not imply that $f_{n} \to f$.

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Neither implies the other. Suppose $(f,f_n)$ is i.i.d. $N(0,1)$. Then $\nu (f_n \in A)=\nu(f \in A)$ for all $A$ but $(f_n)$ deos not converge a.s..

Let $f_n=\frac 1 n$ and $f=0$. Then $f_n \to f$ a.s. but $\nu \{f \in 0\}=1\neq 0=\lim \nu(f_n \in \{0\})$.

Howvever, if $\nu (f_n \in A) \to \nu (f \in A)$ for all $A$ then there exists r.v.'s $g,g_n$ such that $g_n \to g$ a.s., $\nu (f_n \in A)=\nu (g_n \in A), \nu (f \in A)=\nu (g \in A)$. This is by Skorokhod Theorem.

Which can be found: https://encyclopediaofmath.org/wiki/Skorokhod_theorem

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  • $\begingroup$ Thanks. It might be a little late to add clarification but I guess my main question was making sense of the wikipedia article's claim that setwise convergence is the same as probabilists' convergence strongly==convergence almost surely. $\endgroup$
    – kara890
    Aug 5 at 23:20
  • $\begingroup$ @kara890 If Wikipedia says that the two are equivalent then it is simply wrong as my examples show. There is no connection between these two notions of strong convergence. $\endgroup$ Aug 5 at 23:24
  • $\begingroup$ Well, your example says that my interpretation is wrong. But it could be that my interpretation was a wrong guess. $\endgroup$
    – kara890
    Aug 5 at 23:38

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