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Made CDF continuous, otherwise the uniform distribution is false in general.
cgmil
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For a random sequence $X_0, X_1, X_2, \ldots$ and $F_n$ the empirical CDF, does $F_n(X_0)$ converge to a uniform random variable?

Let $X_0, X_1, X_2, \ldots$ be a sequence of i.i.d. real-valued random variables on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$ with continuous CDF $F(x)$ and define a sequence of empirical CDFs $F_n(x) = \frac{1}{n} \sum_{i = 1}^n \mathbf{1}_{\{X_i \leq x\}}(x)$ ($\mathbf{1}_A$ is an indicator function). It is well known that

$$\sup_{x \in \mathbb{R}} \left|F_n(x) - F(x)\right| \to 0 \text{ a.s.}$$

(This is the fundamental theorem of mathematical statistics, or FTMS, supposedly.) Are we then safe to say that

$$F_n(X_0) \to U \text{ in law}$$

where $U$ follows a standard uniform distribution? Or could we make the even stronger claim that we can redefine the random variables into a probability space such that the convergence happens almost surely?

It seems like this would be the case. FTMS suggests we can write $F_n(X_0) = F(X_0) + o(1)$ (holding almost surely) and the distribution of $F(X_0)$ is a standard uniform distribution. While $F_n$ is affected by the asyumptotics, $X_0$ is not. The probability $X_0$ takes a value that is not covered by the convergence is zerom thanks to FTMS being a uniform convergence result. Is it that easy? Am I overthinking this?

cgmil
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