# If $\int_a^x f(t) \mathrm d t=0$ for all $x \in[a, b]$, then $f=0$ $\lambda$-a.e.

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.

• looks good to me Dec 29, 2022 at 14:49
• @Yanko Thank you so much for your verification! Dec 29, 2022 at 16:43