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 asymptotics, $$X_0$$ is not. The probability $$X_0$$ takes a value that is not covered by the convergence is zero thanks to FTMS being a uniform convergence result. Is it that easy? Am I overthinking this?

• I don't see why $F_{X_0}(X_0)$ would be uniformly distributed for an arbitrary rv $X_0$. One obvious problem is that $F_{X_0}$ need not take all values in $[0,1]$. Apr 23, 2022 at 17:18
• @ChristianRemling You are right. CDF should be continuous at least. I fixed the post. Apr 23, 2022 at 19:15

$$\newcommand\Om\Omega\newcommand\om\omega\newcommand\R{\mathbb R}$$This is indeed straightforward. Spelling out $$\sup_{x\in\R}|F_n(x)-F(x)|\to0 \text{ a.s.},$$ we see that for some subset $$N$$ of $$\Om$$ of outer probability $$0$$ and all $$\om\in\Om_0:=\Om\setminus N$$ we have $$\sup_{x\in\R}\Big|\frac1n\sum_{j=1}^n 1(X_j(\om)\le x)-F(x)\Big|\to0$$ and hence $$\frac1n\sum_{j=1}^n 1(X_j(\om)\le X_0(\om))\to F(X_0(\om)).$$
So, indeed we have $$F_n(X_0)\to F(X_0)$$ a.s. and hence in law. Also, if $$F$$ is continuous, then the random variable $$F(X_0)$$ has the standard uniform distribution.