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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?

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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.

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  • $\begingroup$ Thanks. I don't understand how it holds under the last condition. Can you elaborate on your remark, please? $\endgroup$ – master_goon Feb 21 '18 at 20:12
  • $\begingroup$ @master_goon because then for any $\delta>0$ it holds that $|f(X_n)-f_n(X_n)| \leq \delta$ for all but finitely many $n$. And we know that the limit of cesaro means does not depend on the omission of any finite number of terms. $\endgroup$ – Shalop Feb 22 '18 at 1:52
  • $\begingroup$ So if my understanding is correct, one can replace $\|(f_n -f)\cdot 1_{[-\epsilon n , \epsilon n]}\|_{\infty} \to 0$ with $\|(f_n -f)\cdot 1_{\mathbf{B}_n}\|_{\infty} \to 0$ where $\mathbf{B}_n$ is an epsilon ball in $\mathbb{R}^d$ with radius $n$ centered at $0$? $\endgroup$ – master_goon Feb 23 '18 at 12:47
  • $\begingroup$ @master_goon you mean a ball of radius $\epsilon n$, which is true. $\endgroup$ – Shalop Feb 23 '18 at 16:14

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