Strong law of large numbers and uniform convergence on compact sets Let $f:\mathbb{R}^d\to\mathbb{R}$ be a continuous function and let $(X_i)_{i\in\mathbb{N}}$ be idependent and identical copies of the random variable $X$. Suppose all these random variabes are integrable. I know from the strong law of large numbers that
$$
\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f(X_i) = E[f(X)]
$$
with probability one. Suppose we have a sequence of continuous functions $(f_i)_{i\in\mathbb{N}}$ that converges to $f$ uniformly on compact sets. Does it hold that
$$
\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f_n(X_i) = E[f(X)]
$$
with probability one?
I know that if the convergence to $f$ is uniform, then the above is probably true. I'm curious about if it also holds for uniform convergence on compact sets.  
Edit: My thoughts.
The following is to show that if the convergence to $f$ is uniform, then the above is true.
Define
$$
a_n := \sup_{x\in\mathbb{R}^n} |f(x) - f_n(x)|. 
$$
Observe that
$$
\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f(X_i) - f_n(X_i) + f_n(X_i) = E[f(X)]
$$
$$
\implies \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f_n(X_i) - \lim_{n\to\infty} \frac{1}{n}\sum_{i=1}^n f(X_i) + f_n(X_i) = E[f(X)].
$$
Therefore, if we have uniform convergence then $\lim_{n\to\infty} a_n = 0 \implies$ we have the desired result by using Cesaro means to show that the last sum on the left hand side of the equation goes to zero.
Does a similar or better argument exist for when there is uniform convergence on compact sets?
 A: The result is false under the assumption of uniform convergence on compact sets.
Indeed, let the $X_i$ be i.i.d. exponential of rate 1, so $P(X_i>x) = e^{-x}$ for $x \geq 0$.
For $n \geq 1$, we define $f_n(x):= |x|$ for $|x| \leq \frac{1}{2}\log n$, we define $f_n(x) = 2n$ for $|x| \geq \log n$, and we define $f_n$ to be linearly interpolated in between those values. It is clear that the $f_n(x)$ converge to $|x|$ uniformly on compact sets. Moreover, we have that $$\sum_{n=1}^{\infty}P(f_n(X_n) = 2n) = \sum_{n=1}^{\infty} P(X_n\geq \log n) = \sum_{n=1}^{\infty} \frac{1}{n}=+\infty.$$In particular, the second Borel-Cantelli-lemma says that $f_n(X_n)=2n$ infinitely often, a.s. Therefore, $$\limsup_{n \to \infty} \frac{1}{n} \sum_{i=1}^n f_n(X_i) \geq 2 >1 = E|X|.$$Remark: In order for your statement to be true, you need a stronger convergence statement, for instance there exists some $\epsilon>0$ for which $\|(f_n -f)\cdot 1_{[-\epsilon n , \epsilon n]}\|_{\infty} \to 0$. Indeed, by the SLLN (or Borel-Cantelli), we know that $|X_n| \leq \epsilon n$ for all but finitely many $n$, a.s.
