# Glivenko-Cantelli Theorem

The Glivenko-Cantelli Theorem (http://en.wikipedia.org/wiki/Glivenko%E2%80%93Cantelli_theorem) states that if $F$ is a distribution function, $X_1,\dots,X_n \sim F$, and $\hat{F}_n$ is the empirical distribution function, then $$\sup_{x \in \mathbb{R}} \lvert \hat{F}_n(x) - F(x) \rvert \xrightarrow{a.s.} 0 . \tag{1}$$

How does this differ from simply stating the following? $$\hat{F}_n(x) \xrightarrow{a.s.} F(x) \tag{2}$$

Using the definition of convergence almost surely, from (1): \begin{align*} \mathbb{P} \left( \lim_{n\rightarrow\infty} \left\lvert \sup_{x \in \mathbb{R}} \lvert \hat{F}_n(x) - F(x) \rvert \right\rvert = 0 \right) &= 1 \\ \mathbb{P} \left( \lim_{n\rightarrow\infty} \sup_{x \in \mathbb{R}} \lvert \hat{F}_n(x) - F(x) \rvert = 0 \right) &= 1 \tag{1a} \end{align*}

Using the definition of convergence almost surely, from (2): \begin{align*} \mathbb{P} \left( \lim_{n\rightarrow\infty} \left\lvert \hat{F}_n(x) - F(x) \right\rvert = 0 \right) = 1 \tag{2a} \end{align*}

To me, it seems that (1a) and (2a) are equivalent statements because of the least upper bound, and thus (1) and (2) are equivalent statements. But I have a feeling that I'm missing a subtle difference since otherwise I would think the Theorem would just be stated the simpler way (Equation (2)).

The differerence is in uniform convergence. 1 is saying that for all x there is a single n such that error is less than epsilon ( uniform). The other one is saying for each x there is a large enough n that error is less than epsilon(pointwise). An example from wikipedia of pointwise but not uniform convergence is $f_n (x)=x^n$ on $0\le x\le 1$. You need larger and larger n as you get closer to $x=1$.
• Hmm. Did you really mean for that to end with "closer to $x=0$"? I'd have thought that the need for larger $n$ was near $x=1$. – Glen_b -Reinstate Monica Sep 22 '14 at 22:04