Are strong convergence of measure and almost sure convergence of a random variable related? I am studying measure theory and probability I am wondering if and how these two concepts are related.
A sequence of random variables $X_n$ defined on some probability space $(\Omega,\mathscr{A},\mu)$ converges almost surely to $X$ if $\mu\left(\{\lim_{n\rightarrow\infty}X_n=X\}\right)=1.$
A sequence of measures $\mu_n$ converges strongly to $\mu$ if $\lim_{n\rightarrow\infty}\mu_n(A)=\mu(A)$ for all measurable $A$. So $\mu$ and $\mu_n$ are defined on the same measure space $(\Omega,\mathscr{A})$.
Now, is there anyway that these two types of convergence are related?
For example, if we have $X_1,\ldots X_n,\ldots$ are $iid$ standard Gaussian, then the strong law of large numbers says that $\mu(\{\lim_{n\rightarrow\infty}\bar{X}_n=0\})=1$, where I believe $\mu$ can be written as a measure on the space $(\mathbb{R}^\infty,\mathscr{B}(\mathbb{R}^\infty))$. The space on which $\mu$ is defined is one point of confusion.
The sample mean for fixed $n$ also has a Gaussian law with variance $1/n$. We can call this measure $\mu_n$, and it can be defined on the space $(\mathbb{R},\mathscr{B}(\mathbb{R}))$.
Is $\mu_n$ a valid measure on the space of the whole sequence $(\mathbb{R}^\infty,\mathscr{B}(\mathbb{R}^\infty))$, even though it only concerns the first $n$ random variables? Can I set this problem up in a way that says $\mu_n$ converges strongly to $\mu$ ? where $\mu$ is degenerate at 0?
 A: There was such a question: "Is there anyway that these two types of convergence are related?"
Suppose we have $X_n \to X$ almost surely.
Put $P_n (A) = P(X_n \in A)$  and $P(A) = P(X \in A)$ for all Borel set $A$. Then $P_n$, $n \ge 1$ and $P$ are probability measures on real line.
$X_n$ converges to $X$ almost surely, hence $X_n$ converges to $X$ in distribution, and a sequence of measures $P_n$ converges weakly  to the measure $P$. According to Portmanteau Theorem (see Billingsley, Convergence of probability measures, 1999, 2nd edition, pp.15-16) it follows that $P_n(A) \to P(A)$ for all Borel sets $A$ such that $P( \partial A) = 0$. The next example shows that condition  $P_n(A) \to P(A)$ holds not for all Borel sets $A$.
Consider a probability space $(\Omega, \mathcal{F}, \mu)$ where $\Omega = [0,1]$, $\mathcal{F}$ is sigma-algebra of Borel sets of $\Omega$ and $\mu$ is a standard Lebesgue measure, that is $\mu([a,b]) = b-a$. Put $\xi_n(\omega) = n \omega $ for $\omega \in [0, \frac{1}n ]$ and $\xi_n(\omega) = 0$ otherwise. In this case $P_n \sim U[0, \frac{1}{n}]$, $\xi_n \to X \equiv 0$, $P$ is Dirac measure at zero.
For $A = [ - 1, 0] $ we have $P_n(A) = 0 \nrightarrow P(A) = 1$.
As a result we get the next statement. If $X_n$ converges to $X$ almost surely then for corresponding measures $P_n$ and $P$ we have $P_n(A) \to P(A)$ for all Borel sets $A$ such that $P( \partial A) = 0$ but not for all $A$.
Now suppose we have convergence $P_n(A) \to P(A)$ for all $A$ from some sigma-algebra. If these measures are not probability measures of if these measures are not measures on $\mathbb{R}$ with Borel sigma-algebra, then there is no direct relationship between them and any random variables. Suppose that $P_n(A)$, $P(A)$ are probability measures on $\mathbb{R}$ with Borel sigma-algebra and $P_n(A) \to P(A)$ for all Borel sets $A$. Hence $P_n$ converges to $P$ weakly (Portmanteau Theorem). According to the Skorokhod's representation theorem we can construct random variables $X_n$, $X$ such that $X_n$ has distribution $P_n$, $X$ has distribution $P$, $X_n \to X$ almost surely and moreover $X_n$ and $X$ are defined of probability space $(\Omega, \mathcal{F}, \mu)$, where $\Omega = [0,1]$, $\mathcal{F}$ consists of Borel sets of $\Omega$ and $\mu$ is a standard Lebesgue measure, that is $\mu([a,b]) = b-a$.
As a result we get the next statement. If $P_n(A)$, $P(A)$ are probability measures on $\mathbb{R}$ with Borel sigma-algebra and $P_n(A) \to P(A)$ for all Borel set $A$ then there are random variables $X_n, X$ on a "simple" probability space such that $X_n$ has distribution $P_n$, $X$ has distribution $P$ and $X_n \to X$ almost surely.
