a.s. convergence and Glivenko-Cantelli theorem Assume that we have a sample $X_1, X_2, \dots, X_n$ is a sample from, for example, Poisson distribution with probability mass function $p_{i}$ and cdf $\mathbb{F}$. Then, for $i\in\mathbb{N}$ let
$$
\bar{p}_{n,i} = \frac{\sum_{j=1}^{n} 1 \{ X_j = i \}}{n}
$$
is empirical estimator.
The Glivenko-Cantelli theorem says:
$$
\|\bar{\mathbb{F}}_n - F\|_\infty \overset{\text{a.s.}}{\to} 0
$$
where $\bar{\mathbb{F}}_n$ is empirical cdf.
The queston: assume that $\tilde{p}_n$ is another strongly consistent estimator of $p$, i.e. $\tilde{p}_{n}$ converges a.s. to $p$ in $l_{2}$ norm. Note, in $l_{2}$, not point-wise.
Is then
$$
\|\tilde{\mathbb{F}}_n - F\|_\infty \overset{\text{a.s.}}{\to} 0
$$
where $\tilde{\mathbb{F}}_n$ is cdf for the estimator $\tilde{p}_n$?
 A: 
It really depends on how you define $\tilde{{p}_{n,i}}$, for example if you let $$\tilde{{p}_{n,i}} =\overline{{p}_{n,i}}+\frac{1}{n}$$
Then clearly, $ ( \tilde{{p}_{n,i}} ,n) $ is clearly a strongly consistent estimator of $p_i$, however 
$$\tilde{\mathbb{F}}_n( \infty)-F(\infty)= \infty$$
for all $n$, which means $\| \tilde{\mathbb{F}}_n-F\|_{\infty}$ can't converge to $0$.

So we'll be more specific on your "strong consistency" by giving to two following conditions:

*

*$\tilde{p_{n,i}}$ is a strongly consistent estimator of $p_i$

*$\tilde{\mathbb{F}}_n(\infty)= \sum_{i \in \mathbb{N}}\tilde{p_{n,i}}$ converges to $1$ almost surely

So following the same argument in Glivenko-Cantelli theorem's proof, we see that for all $m$, we have:
$$\| \tilde{\mathbb{F}}_n-F\|_{\infty} \le \max_{0 \le i \le m} |\tilde{\mathbb{F}}_n(m)-F(m)|+|F(\infty)-F(m)|+|\tilde{\mathbb{F}}_n(\infty)-\tilde{\mathbb{F}}_n(m)| $$
Thus by the assumed strong consistency, we imply that
$$\limsup_n \| \tilde{\mathbb{F}}_n-F\|_{\infty}  \le 2|1-F(m)|$$
for all $m$, or
$$\limsup_n \| \tilde{\mathbb{F}}_n-F\|_{\infty} =0$$
$\square$
