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.

  • 1
    $\begingroup$ looks good to me $\endgroup$
    – Yanko
    Dec 29, 2022 at 14:49
  • $\begingroup$ @Yanko Thank you so much for your verification! $\endgroup$
    – Analyst
    Dec 29, 2022 at 16:43


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