Prove that $\,f=0$ almost everywhere. 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 $[0,1]$.
 A: Fact. For every $\varepsilon>0$, there exists a 
$\delta>0$, such that $m(E)<\delta$ implies that $\int_E|f|dx<\varepsilon$.
This is due to the fact that $f$ is integrable.
We have that
$$
\{x: f(x)\ne 0\}=\bigcup_{k\in Z} \{x: f(x)\in (2^k,2^{k+1})\}\cup 
\bigcup_{k\in Z} \{x: f(x)\in (-2^{k+1},-2^{k})\}=\bigcup_{k\in Z}A^+_k\cup\bigcup_{k\in Z}A^-_k.
$$
It suffices to show that
$$
m(A_k^+)=m(A_k^-)=0, \quad\text{for all}\,\,\, k\in \mathbb Z.
$$
Assume that $m(A_k^+)=a>0$. For every $\varepsilon>0$, there exists an open set $U$, such that $A_k^+\subset U$ and $m(U\smallsetminus A_k^+)<\varepsilon$. Then
$$
\int_U f\,dx=\int_{A_k^+}f\,dx+\int_{U\smallsetminus A_k^+}f\,dx.
$$
Clearly, $\int_{A_k^+}f\,dx\ge 2^ka$. Using the Fact above, we can choose $\varepsilon$ small enough, so that $\int_{U\smallsetminus A_k^+}|f|<2^{k-1}a$, and therefore
$$
\int_{U\smallsetminus A_k^+}f\,dx\ge-2^{k-1}a,
$$
and hence $\int_U f\,dx\ge2^{k-1}a.$ But, as $U$ is open it can be written as a union of disjoint open intervals: $U=\cup_{n\in\mathbb N}I_n$, and
$$
0<\int_U f\,dx=\sum_{n\in\mathbb N}\int_{I_n}f\,dx,
$$ 
which means that for some interval $I_n=(c,d)$ we should have
$$
\int_c^d f\,dx>0.
$$
Using now the assumption, for every $n\in\mathbb N$, we have
$$
\int_c^d f\,dx=\sum_{k=1}^n\int_{c+\frac{(k-1)(d-c)}{n}}^{c+\frac{k(d-c)}{n}} f\,dx\le n\cdot
\left(\frac{d-c}{n}\right)^2=\frac{(d-c)^2}{n},
$$
which of course implies that $\int_c^d f\,dx=0$, which is a contradiction. Thus $f=0$ a.e.
A: The Lebesgue differentiation theorem states that if $f$ is integrable, then $f(x) = \lim_{\epsilon \downarrow 0} {1 \over 2\epsilon} \int_{x-\epsilon}^{x+\epsilon} f(t) dt$ for ae. [$m$] $x \in [0,1]$. Since $| {1 \over 2\epsilon} \int_{x-. \epsilon}^{x+\epsilon} f(t) dt | \le { (2 \epsilon)^2\over 2 \epsilon} $, we see that $f(x) = 0$ ae. [$m$] $x \in [0,1]$.
A: 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.

