Bounded convergence for expectation of random variables [closed]

Question

I have a random variable $X$ defined on $(0,\infty)$. For each $n\in \mathbb N$, define $X_n = X \mathbf{1}_{0 < X \leq C_n}$, where $C_n$ is a monotonically increasing sequence of positive numbers such that $C_n\to\infty$.

I want to evaluate the limint of an expectation $$ \lim_{n\to\infty} E[f_n(X_n)], $$ where $\{f_n\}_{n\in\mathbb N}$ are a sequence of measurable functions defined on $(0,\infty)$. $f_n(x)$ are bounded by $M$ for all $x$, and $f_n(X_n)$ converges to a constant $c \leq M$ in probability.

In this case, is this true? $$ \lim_{n\to\infty} E[f_n(X_n)] = c. $$

If so, how should I show this?

Answer 1

Yes, it's true. You only need basic facts about convergence in distribution (of real rvs). Both can be e.g. be found in Billingsley's book "Convergence of Probability Measures".

Let $(Y_n)$ be a sequence of real random variables, $Y$ be a real rv.

(1) $Y_n\longrightarrow Y$ in probability implies that $Y_n$ converges to $Y$ in distribution

(2) if $(Y_n)$ is uniformly integrable and converges in distribution to $Y$, then $Y$ is integrable and $\mathbb{E}(Y_n)\longrightarrow \mathbb{E}(Y)$.

Now let $Y_n=f_n(X_n)$. Then $Y_n$ converges to $Y:=c$ in distribution, and since the $Y_n$ are uniformly bounded they are uniformly integrable. Thus $$\mathbb{E}(Y_n)\longrightarrow \mathbb{E}(Y)=c$$