Let $f:\mathbb{R}\rightarrow\mathbb{R^*}$ be a continuous function (when $\mathbb{R^*}=[-\infty,\infty]=2$ points compactification of $\mathbb{R}$, this is not so much necessary for the question).
Let $g(x)=T_x^b(f)$ be the total variation of $f\mid_{[x,b]}$, $g$ is monotonic function, hence $g$ is differentiable almost everywhere, and in particular g is continuous almost everywhere, which means $g$ is preserving convergence almost everywhere, but is $g$ preserving convergence $\textbf{everywhere}?$
But because $f$ is uniformly continuous (continuous on a close interval),
which means $$\forall\epsilon>0 \exists\delta>0:\mid x-y\mid<\delta\Rightarrow\mid f(x)-f(y)\mid<\epsilon$$ and because $$T_x^b(f)=\underset{P}{sup}\Sigma_{i=0}^{n_P}\mid f(x_i)-f(x_{i-1})\mid$$ when $$P_{n_P}:=\{x=t_0< t_1<...<t_{n_P}=b\}$$
It led me to think that $g$ is preserving convergence $\textbf{everywhere}.$
My attempt was to fix a small $\epsilon$, and build a partition $P$ such that the mesh of $P$ is the $\delta$ that derives from this $\epsilon$, but it is not enough to place in the definition of $T_x^b(f)$.
If $g$ is not always preserving convergence for $f\in C([a,b])$, is it preserving convergence for $f\in AC([a,b])?$
Thanks in advance.
