In solving exercise 5.30.4 in Brezis' Functional Analysis, I have come across below lemma
Let $f \in L^1(0, 1)$ such that $\int_0^t f(s) \, \mathrm d s = 0$ for a.e. $t \in (0, 1)$. Then $f=0$ a.e.
Could you have a check on my attempt?
Is there another approach that sidesteps Lebesgue’s differentiation theorem?
By dominated convergence theorem, we have $$ \int_0^t f(s) \, \mathrm d s = \lim_{h \to t} \int_0^h f(s) \, \mathrm d s \quad \forall t \in (0, 1). $$
It follows that $\int_0^t f(s) \, \mathrm d s = 0$ for all $t \in (0, 1)$.
Lebesgue’s differentiation theorem Let $\Omega$ be an open subset of $\mathbb R^n$ and $f \in L_1(\Omega)$. Then for a.e. $x \in \Omega$, we have $$ f(x) = \lim _{r \searrow 0} \frac{1}{|B(x, r)|} \int_{B(x, r)} f(y) \mathrm d y. $$ Here $B(x, r)$ is the open ball of radius $r$ centred at $x$, and $|E|$ denotes the $n$-dimensional Lebesgue measure of $E \subset \mathbb R^n$.
For any $x\in (0, 1)$ and $r>0$ such that $B(x, r) \subset (0, 1)$, we have $$ \int_{B(x, r)} f(y) \mathrm d y = \int_{x-r}^{x+r} f(y) \mathrm d y = \int_0^{x+r} f(y) \mathrm d y - \int_0^{x-r} f(y) \mathrm d y = 0. $$
The claim then follows from Lebesgue’s differentiation theorem.
