If $f\in L^2[0,1]$ such that $\int_0^x f(t)\ dt=0$ for all $x\in [0,1]$. Prove that $f=0$ in $L^2[0,1]$

Question

Let $f\in L^2[0,1]$ such that $\int_0^x f(t)\ dt=0$ for all $x\in [0,1]$. Prove that $f=0$

I know that $L^2[0,1]\subset L^1[0,1]$ and for $f\in L^2[0,1]$, $f$ is zero iff $f= 0$ almost everywhere.

Suppose $f\ne 0$ i.e. there is a measurable set $A$ with $\mu(A)>0$ and $|f|>0$ on $A$. Then $\int_A |f|>0$. But the thing is $\int_A f$ could be zero. How can I show that this $A$ will lead to a contradiction. I can prove the statement if $f$ is continuous.

Can anyone help me in this regard?

Answer 1

Since as you noted $f \in L^2([0,1]) \subset L^1([0,1])$ if you define $$F : [0,1] \rightarrow \Bbb R$$ $$x \mapsto \int_0 ^x f(t)dt$$ Then $$\exists F’(x) = f(x) \:\text{for}\: \mathcal{L}^1 \:\text{a.e.}\: x \in [0,1] \tag{1}$$ but since $F \equiv 0$ then also $f(x) = 0$ for $\mathcal{L}^1 \:\text{a.e.}\: x \in [0,1]$

$(1)$ is a consequence of the Lebesgue differentiation theorem, as suggested in the comment