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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 $\mathbb{R}$.

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    $\begingroup$ Both functions are càdlàg hence the supremum is also the supremum over $\mathbb Q$. $\endgroup$
    – Did
    Jun 20, 2015 at 13:39
  • $\begingroup$ Disregard, I'm tired :) $\endgroup$
    – Math1000
    Jun 20, 2015 at 14:00
  • $\begingroup$ @Did Thanks for the hint, I'll try to work out the details. $\endgroup$
    – simonzack
    Jun 20, 2015 at 14:05

1 Answer 1

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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.)

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  • $\begingroup$ 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? $\endgroup$
    – simonzack
    Jun 20, 2015 at 14:11
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    $\begingroup$ 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. $\endgroup$
    – user21467
    Jun 20, 2015 at 14:17
  • $\begingroup$ I was going to delete this as not actually a correct answer, but maybe the error will be instructive to someone, so I think I'll leave it. $\endgroup$
    – user21467
    Jun 20, 2015 at 15:05

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