Let $X$ be a interval on $\mathbb{R}$ and $g\in C^1(X)$, show that, if $g$ is a function of bounded variation, then $\forall g\in C^1(X)$ we have:
$$\int_{X}^{}|g'(x)|\,dx\,\,=\,\,\sup\{\sum_{j=1}^{+\infty}|g(x_j)-g(x_{j-1})|:(x_j)_{j\in\mathbb{N}\,\,\,\,\text{monotonic sequence of}}\,\,\,X\}$$
Does anyone know how to prove it?
It is clear that RHS $\leq$LHS. For the other way let $\epsilon >0$ and choose $\delta$ such that $|x-y| <\delta$ implies $|g'(x)-g'(y)| <\epsilon $. [ Possible because $g'$ is uniformly continuous]. Consider a partition $\{x_i\}$ of $X$ with $|x_{i+1}-x_i|<\delta$ for all $i$. Now write $\sum |g(x_{i+1})-g(x_i)|$ as $\sum |x_{i+1}-x_i||g'(t_i)|$ for some $t_i$ between $x_{i}$ and $x_{i+1}$ Now $\int_X |g'(t)|\, dt=\sum \int_{x_i}^{x_{i+1}} |g'(t)|\, dt$. From this show that $|\int_X |g'(t)|\, dt-\sum \int_{x_i}^{x_{i+1}} |g'(t_i)|\, dt| < \epsilon$. This gives $\int_X |g'(t)|\, dt< \epsilon + \sum |g(x_{i+1})-g(x_i)|$. Conclude the proof from this.