$\lim_{N\to\infty} \frac{1}{N} \sum_{i=1}^N f(X_i)=\mathbb{E}[f(x)]$

Question

Let $X_i$ be $i.i.d$ integrable random variables with bounded variance and $f$ be a continuous function with compact support. I want to prove $$ \lim_{N\to\infty} \frac{1}{N} \sum_{i=1}^N f(X_i)=\mathbb{E}[f(X)] $$ where the expectation is with respect to a random variable X with the same distribution as $X_i$ in the sequence. Intuitively, the LHS is the empirical average of $f(X_i)$ which will approach to the mean as $N \to \infty$ by SLLN. But I don't know how to use the continuity of $f$ here. Can someone help? Thanks in advance!

Answer 1

This follows immediately from SLLN's since $(f(X_i))$ is also i.i.d. Measurability and boundedness is enough for this; continuity is not needed. [Convergence holds in the almost sure sense].

Answer 2

By chebyshev inequality, $P(|\frac{1}{N} \sum_i f(X_i)-E(f(X))|^2 \geq \frac{1}{N}) \leq \frac{Var(\frac{1}{N} \sum_i f(x_i))}{\frac{1}{N}} = \frac{Var(f(X))}{N} \rightarrow 0 \ as \ N \rightarrow \infty.$

where we used the fact that $Var(f(X))$ is finite. This is for convergence in probability. This is true as long as the set $S=\{z: | z-E(f(X))|^2 \geq \frac{1}{N}\}$ is measurable w.r.t random variable $Z = \frac{1}{N} \sum_i f(X_i)$.