Can we get weakly convergence from the above Glivenko-Cantelli theorem

analysisprobability

I am reading this result:
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Here the empirical distribution function $F_n(x)=\frac{1}{n}\sum I(X_i\le x)$ and $\mu_n=\frac{1}{n}\sum\delta_{X_i}$.

Question: If I want to prove that $\mu_n$ converges to $\mu$ weakly. Can we get this conclusion from the above Glivenko-Cantelli theorem?

Why do we need almost surely $F_n(\cdot)(\omega)\to F$ uniformly on $R$ rather than
$$
\mbox{ almost surely } F_n(x)(\omega)\to F
$$
for each $x\in R$ by SLLN?

Best Answer

Yes, you can since GC implies pointwise convergence of the empirical measures to the limiting measure..

Uniform convergence is a stronger condition than normal convergence:

Uniform Convergence $\implies$ Pointwise Convergence

Therefore $F_n(\cdot)(\omega)\to F$ uniformly a.s $ \implies F_n(x)(\omega)\to F$ for each $x\in R$ a.s.

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