Let $F$ be a CDF and $\hat F$ its empirical analogue. The Glivenko-Cantelli Theorem states that $\Vert \hat F - F\Vert_\infty\rightarrow 0$ almost surely. Can we say anything about the total variation of $\hat F - F$?
Part of the problem is that I don't even know what definition of total variation can be used in this case. I know the definition of bounded pointwise variation for a real-valued function $f$ defined on a bounded interval $[a,b]\subset\mathbb R$: $$V_\text{bpv}(f) := \sup_{P\in\mathcal P}\sum_{i=1}^{n_P}\vert f(x_{i+1}) - f(x_i)\vert,$$ where $\mathcal P$ is the collection of all partitions $P = \{x_1, x_2,\dots, x_{n_P}\}$ of $[a,b]$. There is another (more advanced) definition for a real-valued function defined on an open subset $\Omega\subseteq\mathbb R$: $$V_\text{tcv}(f) := \sup_{h\in\mathcal H} \int_\Omega f \operatorname{div}h,$$ where $\mathcal H$ is the set of all continuously differentiable real-valued functions of compact support contained in $\Omega$ with essential supremum $\leq 1$.
Both definitions are not equivalent. Since the second definition has some advantages (specifically, it allows all of $\mathbb R$), I would go with the second definition. CDFs are real-valued functions defined on $\mathbb R$, so we set $\Omega = \mathbb R$. (Sometimes CDFs are defined as real-valued functions defined on $[-\infty,\infty]$. Does this change anything of the following?) For any $h\in\mathcal H$, we have: $$\int_\mathbb R (\hat F-F)\operatorname{div}h \leq \Vert \hat F - F\Vert_\infty \int_{\mathbb R} \operatorname{div}h = \Vert \hat F - F\Vert_\infty\int_{\mathbb R} h'$$ with $h'$ denoting the derivative of $h$. Let $K$ denote the support of $h$. Then, by definition $h(x) = 0$ for all $x\in\mathbb R\setminus K$, and $h(x)\neq 0$ for $x\in K$. That is, $h'(x) = 0$ for all $x\in \mathbb R\setminus K$ and $h'(x)\neq 0$ for all $x\in K$. Therefore, $$R_h:=\int_\mathbb R h' = \int_K h'.$$ By definition, $h'$ is continuous. So we have a continuous function on a compact set $K$, which means that $h'$ must be bounded. Therefore, there is a $C\geq 0$ such that $\vert h'\vert \leq C$, and since $K$ is compact, it has finite Lebesgue measure. As a result, $R_h$ is bounded. Finally, the Glivenko-Cantelli Theorem implies $V_\text{tcv}(\hat F - F)\rightarrow 0$ almost surely.
Is my reasoning correct? I never worked with total variation before, so I may overlooked something.
Edit: I think that my approach has the following problems:
Derivative at the boundary: I claimed that $h'(x) = 0$ for all $x\in\mathbb R\setminus K$ and $h(x) \neq 0$ for all $x\in K$. But is this is really the case? One could think of a function that has very steep slope close to the boundary. Sure, $h'$ is still continuous on all of $\mathbb R$, so there has to be some regularity. However, I feel that my argument is not rigorous enough.
The upper bound depends on $h$ (resp. $h'$): I show that $R_h$ is bounded. But the supremum over all $h\in\mathcal H$ (of this bound) may not be bounded. That is, my ultimate statement may not be true.
I added a bounty to draw more attention to this question.
Context: In my earlier question, I wanted to control $$I:=\int f\,\mathrm dg$$ with $g = \hat F - F$. By the triangle inequality: $$\left\vert\int f\,\mathrm dg\right\vert \leq \int\vert f\vert\,\mathrm dV(g).$$ I know the result holds for $V = V_\text{bpv}$, but I am not sure whether it holds for $V = V_\text{tcv}$, too. Should this be the case, then, with the assumption of a bounded $f$ and all previous steps being correct, I would have not only determined a bound for $I$ but also showed that $I\rightarrow 0$.
