Prove that if $f \in C^{1}([a,b])$, then $V_{a}^{b} \, f = \int_{a}^{b} |f'(x)| dx$

Question

We are allowed to assume that $f(x_j) - f(x_{j-1}) = \int_{x_{j-1}}^{x_j} f'(x) dx.$

So I have shown that given any partition $P$ of the interval $[a,b]$, that $T(f,P) \leq \int_{a}^{b} |f'(x)| \, dx.$ Taking the supremum of $T(f,P)$ over all possible partitions $P$ of $[a,b]$, we get that:

$V_{a}^{b} f \leq \int_{a}^{b} |f'(x)| \, dx.$

I am stuck here on trying to prove equality. I know I haven't explicitly used the fact that $f'$ exists on $[a,b]$ and is continuous on $[a,b]$ but I don't see how that helps me.

Any and all help is very greatly appreciated. Thanks in advance!!!

Answer 1

Here is slightly different proof: For each partition $P = \{a = x_0 < x_1 < \cdots < x_n = b\}$, the mean value theorem shows that

$$ L(|f'|, P) = \sum_{k=1}^{n} \Big( \inf_{[x_{k-1},x_k]} |f'| \Big) (x_k - x_{k-1}) \leq T(f, P) \leq V_{a}^{b}(f), $$

where $L(|f'|,P)$ is the lower Riemann sum of $|f'|$. Since $|f'|$ is continuous, it is Riemann integrable and hence

$$ \int_{a}^{b} |f'(x)| \, dx = \sup_P L(|f'|, P) \leq V_{a}^{b}(f). $$

The other direction is already proved, hence we have the equality.