If the laws of random variables weakly converge to a point mass at $c$, then the random variables converge in probability to $c$. This is question 10.3.1 from Rosenthal's First Look at Rigorous Probability.
Suppose $\mathcal{L}(X_n) \Rightarrow \delta_c$ for some $c \in \mathbb{R}$. Prove that $\{X_n\}$ converges to $c$ in probability.
Here $\mathcal{L}(X_n)$ refers to the distribution of $X_n$.  Is the following proof correct?  Is a simpler proof available?  Is there a simple proof using Skorohod's theorem?
I attempted to prove the contrapositive. Suppose $X_n \not \to c$ in probability.  This means that there is an $\varepsilon > 0$ so that for all $N$, there exists an $m > N$ such that
\begin{equation}
  P(|X_m - c| \geq \varepsilon) > \eta, \text{ for some $\eta > 0$}.
\end{equation}
Consider
\begin{equation}
  B := \{x \in \mathbb{R} : |x - c| \leq \varepsilon\} \in \mathcal{B} \text{ the Borel sets on $\mathbb{R}$}.
\end{equation}
Then $B^c \in \mathcal{B}$ and denoting the distribution of $X_n$ as $\mu_n$ we have
\begin{equation}
  \mu_n(B^c) = P(X_n \in B^c) = P(|X_n - c| \geq \varepsilon) > \eta.
\end{equation}
For all $N$, there exists an $m > N$ such that the cumulative distribution function of $X_m$ is not the same as the cumulative distribution function of $\delta_c$; in particular it's mass changes somewhere at least $\varepsilon$ distance from $c$. Finally because $\delta_c(\{x\}) = 0$ for all $x \neq c$ this completes the proof.
 A: Here's another possible proof that is maybe more direct.  Let $\mu_n$ denote the law of $X_n$.
Fix $\epsilon > 0$.  We want to show $P(|X_n -c| \le \epsilon) \to 1$.  Construct a continuous function $f$ such that:


*

*$0 \le f \le 1$;

*$f = 0$ outside $(c-\epsilon, c+\epsilon)$;

*$f(c) = 1$.


Then we have
$$P(|X_n - c| \le \epsilon) = \int 1_{[c-\epsilon, c+\epsilon]}\,d\mu_n \ge \int f\,d\mu_n \to \int f\,d \delta_c = f(c) = 1.$$
You could also use Skorohod's theorem if you want.  Let $Y_n$ be the sequence of random variables produced by Skorohod, with $Y_n \sim \mu_n$.  Then we have $Y_n \to c$ almost surely and in particular $Y_n \to c$ in probability.  That is, $P(|Y_n - c| \le \epsilon) \to 1$.  But $$P(|Y_n - c| \le \epsilon) = \mu_n([c-\epsilon, c+\epsilon]) = P(|X_n -c| \le \epsilon),$$ so $X_n \to c$ in probability as well.  However, this seems like using too much machinery to me.
A: This question was over a decade ago but I would like to leave an alternative proof, which is probably easier for most people to understand.
As you said in your comment that weak convergence is equivalent to $\mu_n(-\infty, x] \to \mu(-\infty, x]$, so let's write the distribution functions as follows: $F_n(x) = \mu_n(-\infty, x]$ and $F(x)= \mu(-\infty,x]$. Then, the assumption that the laws of random variables converge to a point mass at $c$ implies that $F(x) = I(x \geq c)$ where $I(A)$ is an indicator function that evaluates to $1$ if $A$ is true and 0 otherwise. This also implies that $c-\epsilon$ and $c+\epsilon$ for any $\epsilon>0$ are points of continuity for $F$. Therefore, we can deduce that $F_n(c-\epsilon) \to F(c-\epsilon)=0$ and $F_n(c+\epsilon) \to F(c+\epsilon)=1$.
This in turn gives you $P(-\epsilon < X_n - c < \epsilon) = F_n(c+\epsilon) - F_n(c-\epsilon)$, which tends to 1. Noticing $P(-\epsilon < X_n - c < \epsilon) = P(|X_n-c|<\epsilon)$ completes the proof.
The observation here is that you don't need to rely on the contrapositive but rather derive the probability of the set $|X_n-c|<\epsilon$ that we're interested in for in-probability convergence, directly.
