Let $f$ be a Lebesgue integrable function on $[0,1]$ such that, for any $0 \leq a < b \leq 1$,
$$\biggl|\int^b_a f(x)\,dx\,\biggr| \leq (b-a)^2\,.$$
Prove that $f=0$ almost everywhere.
I would be thankful if somebody tell me whether my attempt is correct or not:
Consider the partition
$$
[0,1]=\left[0,\frac{1}{n}\right) \cup \left[\frac{1}{n},\frac{2}{n}\right)
\cup \cdots \cup \left[\frac{n-1}{n},1\right].
$$
I am going to use the fact that if $g$ is an integrable over a measurable set $E$, $g$ is finite almost everywhere on $E$, i.e. there is $M>0$ such that
$\vert g\rvert<M$ a.e. on E, moreover $$
\bigg|\int_Eg(x)dx\,\bigg|\leq M m(E)
$$
Now using the hypothesis
$$
\biggl|\int^{\frac{1}{n}}_{0} f(x)\,dx\biggr| \leq \biggl( \frac{1}{n} \biggr)^2
\Rightarrow |f(x)| \leq \frac{1}{n} \ \ a.e. \ on \ \biggl[0,\frac{1}{n} \biggr)
$$
$$\biggl|\int^{\frac{2}{n}}_{\frac{1}{n}} f(x)\,dx\biggr| \leq \biggl(\frac{1}{n}\biggr)^2 \Rightarrow |f(x)| \leq \frac{1}{n} \ \ a.e. \ on \ \biggl[\frac{1}{n},\frac{2}{n}\biggr) $$
and similarly
$$\biggl|\int^{1}_{\frac{n-1}{n}} f(x)\,dx\biggr| \leq \biggl(\frac{1}{n}\biggr)^2 \Rightarrow |f(x)| \leq \frac{1}{n} \ \ a.e. \ on \ \biggl[\frac{n-1}{n},1\biggr]$$
Hence,
$$ |f(x)| \leq \frac{1}{n} \ \ a.e. \ on \ [0,1]$$ we can let $n \rightarrow \infty$ which yields that $f=0$ almost everywhere on $[0,1]$.
Best Answer
Define $g(x)=\int_0^x f(t)dt$. Then $g$ satisfies $|g(x)-g(y)|\leq (x-y)^2$. This implies at once that $g$ is continuous, differentiable with $g'(x)=0$. Therefore $g$ is constant and $g(0)=0$ implies that $g(x)=0$.
This means that $f$ is integrable with $\int_I f=0$ for every interval $I \subset [0,1]$ and this extends to $\int_B f=0$ for every Borel measurable set $B \subset [0,1]$. Consider now the sets $A_n = \{ |f| \geq \frac{1}{n}\}$. Then the sets $A_n$ are Lebesgue measurable and we have $\mu(A_n)=\sup\limits_{K \subset A_n, \text{ compact} }\mu(K)=0$ (since compact sets are Borel measurable; $\mu$ is the Lebesgue measure).
Therefore $$\mu(\{f\neq 0\})=\mu(\bigcup_n A_n) =\lim_{n\to \infty} \mu(A_n)=0 $$ so $f=0$ almost everywhere.