# Let $f$ be a bounded uniformly continuous function in $R^1$. Then $X_n\to 0$ in pr. implies $E(f(X_n)) \to f(0)$.

The following is problem 4 from Section 4.2 of "A Course in Probability Theory" by Kai Lai Chung.

Let $f$ be a bounded uniformly continuous function in $R^1$. Then $X_n\to 0$ in pr. implies $E(f(X_n)) \to f(0)$.

I don't understand where the condition uniformly continuous is needed. Where is the flaw in the following argument that uses just $f$ is continuous at 0?

Consider an $\epsilon >0$.

Continuity of $f$ says that there exists $\delta > 0$ such that if $|x| \le \delta$ then $|f(x)-f(0)|<\epsilon/3$

That $f$ is bounded says that there exists $B$ such that $|f|<B$.

Because $X_n\to 0$ in pr. we know that there exists $N$ such that if $n> N$ then $P(|X_n| > \delta) < \epsilon/3B$.

Then for each $n > N$ we have that $$|E(f(X_n))-f(0)|=|\int f(X_n)dP - f(0)| = \left|\int_{|X_n|>\delta} f(X_n) + \int_{|X_n|\le \delta}(f(X_n)-f(0)) - \int_{|X_n|>\delta}f(0)\right|\le \left|\int_{|X_n|>\delta} f(X_n)\right| + \left|\int_{|X_n|\le \delta}(f(X_n)-f(0))\right| + \left|\int_{|X_n|>\delta}f(0)\right|\le \int_{|X_n|>\delta} |f(X_n)| + \int_{|X_n|\le \delta}|f(X_n)-f(0)| + \int_{|X_n|>\delta}|f(0)| \le \int_{|X_n|>\delta} B + \int_{|X_n|\le \delta}\epsilon/3 + \int_{|X_n|>\delta}B \le B*(\epsilon/3B)+\epsilon/3+B*(\epsilon/3B) = \epsilon$$

Let $f$ be a bounded uniformly continuous function in $R^1$. Then $X_n\to X$ in pr. implies $E(f(X_n)) \to E(f(X))$.
• What exactly do you mean by $E(f(X_n)) \to f(X)$ as $E(f(X_n))$ is a number but $f(X)$ is a random variable? Commented May 26, 2013 at 8:30