Approximation of sample mean distribution Suppose that we have an iid sample $X_1,\dots,X_n$ with a distribution function $F$. 
Denote $\bar X_n:=\frac{1}{n}\sum_{i=1}^n X_i$ and $\bar X_n^*:=\frac{1}{n}\sum_{i=1}^n X_i^*,$ where $X_1^*,\dots,X_n^*$ are iid from the empirical distribution function $\hat F$ given the sample $X_1,\dots,X_n$. Does it then follow that 
$$\sup_x|\mathbb P(\sqrt{n}\bar X_n \leq x) - \mathbb P(\sqrt{n}\bar X_n^* \leq x)| \rightarrow 0 \ \ \mathrm{a.s.}$$
 A: Suppose the variance of the distribution $F$ is $\sigma^2<\infty$, and denote $\mu$ the mean of $X_i$. By Lindeberg–Lévy CLT, we have
$$\mathbb{P}\left(\sqrt{n}\bar{X}_n\leq x\right)\rightarrow_d N\left(\mu,\sigma^2\right).$$
Denote $\Phi_{\mu,\sigma}(x)$ the CDF of a normal distributed random variable with mean $\mu$ and variance $\sigma^2$ evaluated at $x$. Because $\Phi_{\mu,\sigma}(x)$ is continuous throughout $\mathbb{R}$, we further have
$$\sup_x\left|\mathbb{{P}}\left(\sqrt{n}\bar{X}_{n}\leq x\right)-\Phi_{\mu,\sigma}\left(x\right)\right|\rightarrow0,\,n\rightarrow\infty.$$
Next, we can show that
$$\mathbb{P}\left(\sqrt{n}\bar{X}_n^*\leq x\right)\rightarrow_d N\left(\mu,\sigma^2\right).$$
Note that $X_1^*,\ldots,X_n^*$ are iid, and
$$E\left(X_{i}^{*}\right)=E\left\{ E\left(X_{i}^{*}\mid X_{1},\ldots,X_{n}\right)\right\} =E\left(\frac{1}{n}\sum_{i=1}^{n}X_{i}\right)=\mu.$$
Similaryly, we have $Var(X_i^*)=\sigma^2$. Then the above convergence follows from the CLT. And the convergence is also uniform.
In the end, your claims holds because
\begin{aligned} & \sup_{x}\left|\mathbb{P}\left(\sqrt{n}\bar{X}_{n}\leq x\right)-\mathbb{P}\left(\sqrt{n}\bar{X}_{n}^{*}\leq x\right)\right|\\
= & \sup_{x}\left|\mathbb{P}\left(\sqrt{n}\bar{X}_{n}\leq x\right)-\Phi_{\mu,\sigma}\left(x\right)+\Phi_{\mu,\sigma}\left(x\right)-\mathbb{P}\left(\sqrt{n}\bar{X}_{n}^{*}\leq x\right)\right|\\
\leq & \sup_{x}\left|\mathbb{P}\left(\sqrt{n}\bar{X}_{n}\leq x\right)-\Phi_{\mu,\sigma}\left(x\right)\right|+\sup_{x}\left|\mathbb{P}\left(\sqrt{n}\bar{X}_{n}^{*}\leq x\right)-\Phi_{\mu,\sigma}\left(x\right)\right|\\
= & 0.
\end{aligned}
