# Weak convergence of measures implying almost sure convergence of random variables

Suppose $$\mu,\mu_n$$ are Borel probability measures on $$\mathbb{R}$$ with $$\mu_n$$ converging weakly to $$\mu$$. I am asked to find some probability space $$(\Omega,\mathcal{F},\mathbb{P})$$ and random variables $$X,X_n$$ such that $$X$$ has law $$\mu$$, $$X_n$$ has law $$\mu_n$$ and $$X_n \to X$$ almost surely as $$n \to \infty$$.

So far I tried to let $$(\Omega,\mathcal{F},\mathbb{P}) = ((0,1),\mathcal{B}_{0,1},\text{Lebesgue})$$ and defined $$X_n$$ as for $$\omega \in (0,1)$$ we let $$X_n(\omega) = \inf\{x \in \mathbb{R}: \omega \in \mu_n((-\infty,x])\}$$. Then $$\mathbb{P}(X_n \in (-\infty,x]) = \mathbb{P}(X_n^{-1}((-\infty,x])) = \mu_n((-\infty,x])$$ so $$X_n$$ has law $$\mu_n$$ and similarly for $$X$$.

However I am not able to prove that $$X_n \to X$$ almost surely. I have tried using contradiction, if $$X_n \not\to X$$ almost surely then we have that $$|X_n-X| \geq 0$$ and this does not converge to 0 almost surely. So the integral of this does not converge to $$0$$ as $$n \to \infty$$. However at this point I get stuck. I can't see where I should bring in the fact that $$\mu_n \to \mu$$ weakly.

How should I proceed?

• This is a special case of Skhorohod Representation Theorem. You can search Wikipedia for this theorem. – Kavi Rama Murthy Sep 22 '18 at 12:42

## 1 Answer

Let $$\mu_n$$, $$\mu$$ be probability measures on $$\mathbb{R}$$ such that $$\mu_n$$ converges weakly to $$\mu$$. The associated cumulative distribution functions $$F_n(x) := \mu_n((-\infty,x]) \qquad F(x) := \mu((-\infty,x])$$ then satisfy $$F_n(x) \to F(x) \tag{1}$$ for any continuity point $$x$$ of $$F$$.

Lemma Denote by \begin{align*} F_n^{-1}(t) &:= \inf\{x \in \mathbb{R}; F_n(x) \geq t\} \\ F^{-1}(t) &:= \inf\{x \in \mathbb{R}; F(x) \geq t\} \end{align*} the generalized inverse functions of $$F_n$$ and $$F$$, respectively. If $$t \in (0,1)$$ is a continuity point of $$F^{-1}$$, then $$\lim_{n \to \infty} F_n^{-1}(t) = F^{-1}(t). \tag{2}$$

Once we have proved the result, we find immediately that the sequence of random variables defined in your question $$X_n(t) = F_n^{-1}(t) \qquad X(t) = F^{-1}(t)$$ satisifies $$X_n(t) \to X(t)$$ for any continuity point $$t$$ of $$F^{-1}$$; as $$F^{-1}$$ has at most countably many discontinuies, this shows $$X_n \to X$$ almost surely.

Proof of the lemma: Let $$t \in (0,1)$$ be a continuity point of $$F^{-1}$$. Since $$F$$ has at most countably many discontinuities, we can find for any $$\epsilon>0$$ a continuity point $$x$$ of $$F$$ such that $$F^{-1}(t)- \epsilon< x < F^{-1}(t).$$ As $$x < F^{-1}(t) \implies F(x) < t$$ and $$\lim_n F_n(x) = F(x)$$, we have $$F_n(x) for large $$n \in \mathbb{N}$$, and the definition of the inverse yields $$x \leq F_n^{-1}(t)$$. Consequently, $$F^{-1}(t) - \epsilon < x \leq F_n^{-1}(t)$$ implying $$F^{-1}(t) \leq \liminf_{n \to \infty} F_n^{-1}(t).$$ It remains to show that $$\limsup_{n \to \infty} F_n^{-1}(t) \leq F^{-1}(t). \tag{3}$$ To this end, we note that for any $$u>t$$ and $$\epsilon>0$$ we can find a continuity point $$y$$ of $$F$$ such that $$F^{-1}(u) < y < F^{-1}(u)+\epsilon.$$ This implies $$F(y) \geq u > t.$$ As $$y$$ is a continuity point of $$F$$, this means that $$F_n(y) \geq t$$ for large $$n \in \mathbb{N}$$. Hence, $$y \geq F_n^{-1}(t)$$, and so $$F^{-1}(u)+\epsilon > y \geq F^{-1}_n(t)$$ for large $$n$$. Thus, $$\limsup_{n \to \infty} F_n^{-1}(t) \leq F^{-1}(u).$$ Letting $$u \downarrow t$$ we find from the (right)continuity of $$F^{-1}$$ at $$t$$ that $$(3)$$ holds.

Reference: Theorem 8.3.2 in S.I. Resnick: A probability path, Birkhäuser 2013.