Let $\lambda$ be the Lebesgue measure on $[a, b]$. We endow $[a, b]$ with the Lebesgue $\sigma$-algebra $\mathcal L([a, b])$. Then $([a, b], \mathcal L([a, b]), \lambda)$ is a complete measure space. I would like to see how Lebesgue integration fits in the framework of Bochner integration by proving below result, i.e.,
Theorem Let $f:[a, b] \rightarrow \mathbb{R}$ be $\lambda$-integrable such that $$ \int_a^x f(t) \mathrm d t=0 \quad \text {for all } x \in[a, b]. $$ Then $f=0$ $\lambda$-a.e. on $[a, b]$.
Could you have a check on my attempt?
Proof It suffices to prove that $$ \int_A f(t) \mathrm d t = 0 \quad \forall A \in \mathcal L([a, b]). $$
We have $\mathcal L([a, b])$ is the completion of the Borel $\sigma$-algebra $\mathcal B([a, b])$ of $[a, b]$ with respect to $\lambda$. So it suffices to prove that $$ \int_A f(t) \mathrm d t = 0 \quad \forall A \in \mathcal B([a, b]). $$
Let $\tau$ be the metric topology of $[a, b]$. Every set in $\tau$ is a countable union of disjoint open intervals. Hence $$ \int_O f(t) \mathrm d t = 0 \quad \forall O \in \tau. $$
Fix $A \in \mathcal B([a, b])$. Because $\lambda$ is outer regular, there is a sequence $(O_n) \subset \tau$ such that $A \subset O_{n+1}\subset O_n$ and that $\lim_n \lambda(O_n) = \lambda(A)$. By dominated convergence theorem, $$ \int_{A} f(t) \mathrm d t = \lim_n \int_{O_n} f(t) \mathrm d t = 0. $$
This completes the proof.
