Consider a function $f\in L^1([a,b])$ and define $F(x):=\int_a^x f(y)dy$. I should prove that $V_a^bF=||f||_{L^1([a,b])}$.
My attempt was to proceed by approximation with a test function $\phi$, but I’m not getting it. Can someone please help me, please?
Consider a partition of $[a,b]$ which splits $F(x)$ into positive and negative parts (continuity of F). Taking limits, this should give you your lower bound. The upper bound is an exercise in using the triangle inequality.
Take any partion $\Pi = \{a = t_0 \leq \ldots \leq t_n = b\}$. Then \begin{align*} V_a^b(F, \Pi) = \sum_{i=1}^n|F(t_i) - F(t_{i-1})| &= \sum_{i=1}^n\Big|\int_{t_{i-1}}^{t_i}f(y)dy \Big| \\ &\overset{*}{\leq} \sum_{i=1}^n\int_{t_{i-1}}^{t_i}|f(y)|dy \\ &= \int_a^b|f(y)|dy \\ &= ||f||_{L^1([a,b])}, \end{align*} where we used Jensen's inequality at the $*$-step. Hence, $V_a^b(F) \leq ||f||_{L^1([a,b])}$. For the lower bound, you can use the fundamental theorem of calculus for Lebesgue integrals. That is, $F$ is differentiable on a set $N$ of Lebesgue measure $1$, and for $x \in N$, $F'(x) = f(x)$. For simplicity, we will assume $F$ to be everywhere differentiable, and $f$ to be Riemann integrable. (You fill in the rest of the details.) Now, take any sequence of partitions $\Pi^n = \{a=t_0^n \leq \ldots \leq t_{k_n}^n = b\}$ such that $\max_{1\leq i \leq k_n^n} |t_{i}^n-t_{i-1}^n| \to 0$ as $n \to \infty$. Then, \begin{align*} V_a^b(F, \Pi^n) = \sum_{i=1}^{k_n}|F(t_i^n) - F(t_{i-1}^n)| &= \sum_{i=1}^{k_n}\frac{|F(t_i^n) - F(t_{i-1}^n)|}{(t_i^n-t_{i-1}^n)}(t_i^n-t_{i-1}^n)\\ &= \sum_{i=1}^{k_n}|f(c_i^n)|(t_{i}^n - t_{i-1}^n) \\ &\to \int_a^b|f(y)|dy \\ &= ||f||_{L^1([a,b])}. \end{align*} as $n \to \infty$. Here, the values $c_i^n$ are determined by the mean value theorem. So indeed, $V_a^b(F) = ||f||_{L^1([a,b])}$.
