# The variation of a function as a function

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<...

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 $$f\in AC([a,b])?$$

Try $$f(x)=xsin(\frac{1}{x})$$ with $$0$$ at $$x=0$$. Then for every $$\epsilon>0$$ $$T_0^\epsilon(f)=\infty$$ but $$T_0^0(f)=0$$.