# Why is the supremum a random variable in the Glivenko–Cantelli theorem

According to wikipedia:

Assume that $X_1,X_2,\dots$ are independent and identically-distributed random variables in $\mathbb{R}$ with common cumulative distribution function $F(x)$. The empirical distribution function for $X_1,\dots,X_n$ is defined by

$$F_n(x)=\frac{1}{n}\sum_{i=1}^n I_{(-\infty,x]}(X_i)$$

where $I_C$ is the indicator function of the set $C$.

...

Theorem

$$\|F_n - F\|_\infty = \sup_{x\in \mathbb{R}} |F_n(x) - F(x)| {\longrightarrow} 0$$ almost surely.

Now why is $\sup_{x\in \mathbb{R}} |F_n(x) - F(x)|$ even a random variable (i.e. that is, it's measurable)? I know the supremum for a countable set of RVs is a random variable, but here it's over an uncountable set $\mathbb{R}$.

• Both functions are càdlàg hence the supremum is also the supremum over $\mathbb Q$.
– Did
Jun 20, 2015 at 13:39
• Disregard, I'm tired :) Jun 20, 2015 at 14:00
• @Did Thanks for the hint, I'll try to work out the details. Jun 20, 2015 at 14:05

Not a probabilist, but I wonder if something like the following would work: $F_n$ and $F$ are increasing, so they're continuous except at countably many points; in each interval where both are continuous, you're dealing with a supremum of continuous functions, which (insert technical details about compactness) is continuous, hence measurable; and then the entire supremum is over only countably many such intervals.
(To clarify (?): I mean to write $\displaystyle\sup_{x\in\mathbb R} = \sup_I \sup_{x\in I}$, where $I$ ranges over the intervals of continuity.)
• Interesting idea but what if, say $F$ is discontinuous at $\mathbb{Q}$ (not sure this is possible or not though)? Then what will the intervals $I$ be? Jun 20, 2015 at 14:11
• Hm, yes, that seems to ruin my idea. It's possible like this: enumerate the rationals somehow as $\mathbb Q = \{q_n\colon n\in\mathbb N\setminus\{0\}\}$ and let $P(X=q_n) = \frac1{2^n}$. Then $F$ is discontinuous exactly at the rationals.