Let $f:[a,b]\to \mathbb R$ absolute continuous, i.e. there is $g\in L^1(a,b)$ s.t. $$f(x)-f(y)=\int_x^y g.$$
Why such function has bounded variation on $[a,b]$ ?
In my solution they say that it's a consequence of the fact that $$\forall \varepsilon >0, \exists \delta >0: \forall E\text{ measurable s.t. }m(E)<\delta \implies \int_E f<\varepsilon .\tag{(*)}$$
But I don't really understand in what is useful. Don't we have the following fact : Let $\pi: a=t_0<t_1<...<t_n=b$ a subdivision.
$$\sum_{i=0}^{n-1}|f(t_{i+1})-f(t_i)|\leq \sum_{i=0}^{n-1}\int_{t_{i}}^{t_{i+1}}|g|=\int_a^b |g|,$$
Therefore, $$\sup_{\{a=t_0<...<t_n=b\} \in \Pi(a,b)}\sum_{i=0}^{n-1}|f(t_{i+1})-f(t_i)|\leq \int_a^b|g|<\infty ,$$
and thus has bounded variation. Isn't it correct ? If yes, in what $(*)$ is relevant ?