Total Variation and indefinite integrals

Suppose $f$ is Lebesgue integrable on $[a,b]$ and $F(x) = \int^x_a f(t) dt$, $x \in [a,b]$. Show that $F$ has bounded variation, and the total variation $T^b_a(F)$ satisfies $$ T^b_a(F) = \int^b_a |f(t)|dt. $$

Now, here is what I have so far. Show $F$ is of BV is simple: let $a=x_0 < x_1 < \dots < x_k=b$ be any partition/subdivision of $[a,b]$. Then $$ \sum^k_{i=1} |F(x_i)-F(x_{i-1})| = \sum^k_{i=1} \left|\int^{x_i}_{x_{i-1}} f(t) dt\right| \leq \sum^k_{i=1} \int^{x_i}_{x_{i-1}} |f(t)| dt = \int^b_a |f(t)| dt. $$ Thus $T^b_a(F) \leq \int^b_a|f(t)| dt < \infty$. Therefore $F$ is of bounded variation.

On the other hand, I also know that $F$ is an indefinite integral iff $F$ is absolutely continuous. This would have also implied $F$ is BV since AC $\Rightarrow$ BV. I also know that if $F$ is BV, then $F'$ exists a.e.(almost everywhere). I also have an old homework question I did which showed $F'=f$ a.e. with the assumptions from the start. Thus all I need is that $\int|F'| \leq T^b_a(F)$.

I think this works: for every $x \in [a,b]$, we know that $F(x) = P^x_a(F) - N^x_a(F) + F(a)$; P and N are the positive and negative variations respectively. Then $\dfrac{d}{dx} [F(x)] = \dfrac{d}{dx}[P^x_a(F)] - \dfrac{d}{dx}[N^x_a(F)]$ a.e. Thus $|F'| \leq \dfrac{d}{dx}[P^x_a(F)] + \dfrac{d}{dx}[N^x_a(F)] = \dfrac{d}{dx}[T^x_a(F)]$ a.e. Finally $$ \int^b_a |F'| \leq \int^b_a \dfrac{d}{dx} T^x_a(F) \leq T^b_a(F). $$ Assuming I can show anything I have stated and didn't show proof of, this looks like it works to me. Any comments would be greatly appreciated.

Solution 1:

I think your proof that $\int|F'| = T^b_a(F)$ works. However, the easier way to go, since $F$ is differentiable, is to apply the mean value theorem to an arbitrary partition.