# Interchange of limit and discontinuous function

I encountered the following statement in some lecture notes and it does not seem right to me. I just wanted to have my thoughts verified or falsified.

$$(X_n)_{n\in\mathbb{N}}$$ is a sequence of random variables and $$X_n \to X$$ almost surely for some r.v. $$X$$. The claim is as follows. $$P\{X_n > b\} \to P\{X > b\}$$ for any $$b \in \mathbb{R}$$.

I have a counter example for this claim. Take a probability space with two elements: $$\omega_1,\omega_2$$. Set $$P\{\omega_1\} = P\{\omega_2\} = 1/2$$.

Define $$X_n(\omega_1) = 1 + \frac{1}{n}$$ and $$X_n(\omega_2) = 0$$. Then define $$X(\omega_1) = 1$$ and $$X(\omega_2) = 0$$ so that the convergence hypothesis holds.

Take $$b = 1$$. $$P\{X_n > b\} = 1/2$$ for every $$n$$ but $$P\{X > b\} = 0$$. So the assertion in the lecture notes seems to be false. The reason is as follows.

While it is correct by the dominated convergence theorem that

$$\lim_{n\to\infty}P\{X_n > b\} = E[ \lim_{n\to\infty} 1_{X_n > b}]$$

it is not true that

$$E[ \lim_{n\to\infty} 1_{X_n > b}] = E[ 1_{\underbrace{\lim_{n\to\infty} X_n}_X > b}]$$ since the indicator function is not continuous.

• Your counterexample looks convincing to me. – Giuseppe Negro Sep 23 '18 at 12:25
• @GiuseppeNegro Thanks. – Calculon Sep 23 '18 at 13:30