For any $n$, let $X_1, …, X_n$ be i.i.d. random variables on the probability space $(\Omega, \mathcal{F}, Pr)$ where $\Omega = \mathbb{R}^d$. Define
$$
\mu_n(A) = \frac{1}{n} \sum_{i=1}^n 1_{\{X_i \in A\}}, \\
\mu(A) = Pr(X_1 \in A).
$$
Consider any $\mathcal{A} \subset \mathcal{F}$, and define
$$
g(X_1, …, X_n) := \sup_{A \in \mathcal{A}} |\mu_n(A) – \mu(A)|
$$
Prove that $g(X_1, …, X_n)$ converges in probability to $0$ implies that $g(X_1, …, X_n)$ converges almost surely to $0$.
p/s: This problem is posed as an exercise in the book "Combinatorial methods in density estimation" by Luc Devroye and Gabor Lugosi (exercise 3.2 with a hint to use the bounded difference inequality). I have tried but could not solve it. Hopefully, someone could help. Thank you.
Edit: As noted by @NickyLevering, I change "iff" to "implies" because it is well-known that a.s. convergence implies convergence in probability (also changed the title to reflect this better). I also set $\Omega = \mathbb{R}^d$ to make it clearer.
Best Answer
Abbreviate $g_n=g(X_1,...,X_n)$ and note, first of all, that for any $A$,
$$ \left\|\sum_{i=1}^n 1_{X_i\in A}-\sum_{i=1}^n 1_{Y_i\in A}\right\|\leq |\{i|X_i\neq Y_i\}| $$ Hence, we can apply the bounded differences inequality with $c_i=\frac{1}{n}$ and get that
$$ Pr(|g_n-\mathbb{E} g_n|\geq t)\leq 2\exp(-2tn) $$ for every $t>0$.
In particular, $$ \sum_{n=1}^{\infty}Pr(|g_n-\mathbb{E} g_n|\geq \frac{1}{\sqrt{n}})\leq 2 \sum_{n=1}^{\infty}\exp(-2\sqrt{n})<\infty, $$ so by Borel Cantelli, $g_n-\mathbb{E} g_n\to 0$ almost surely.
Thus, if $g_n\to 0$ in probability, we want to prove that $\mathbb{E}g_n$ goes to $0$. This is equivalent to proving that every subsequence $g_{n_k}$ has a subsequence $g_{n_{k_j}}$ such that $\mathbb{E} g_{n_{k_j}}\to 0$. This follows since $g_{n_k}\to 0$ in probability and hence, has a subsequence $g_{n_{k_j}}\to 0$ almost surely. Then, since $0\leq g_{n_{k_j}}\leq 1$, we can apply Dominated Convergence to get $\mathbb{E}g_{n_{k_j}}\to 0$.