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?



Bounded convergence for expectation of random variables
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 $$\mathbb{E}(Y\_n)\longrightarrow \mathbb{E}(Y)=c$$


