convergence setwise versus convergence almost surely

Question

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?

Answer 1

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 $f_{n} \to f$.

Answer 2

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