I am currently reviewing old exams to prep for my own upcoming exam. This is a problem I am having trouble with, I'm sure I'm overlooking something simple:
Show that $f=0$ ae on [0,1] if $\int_E f \le m(E)^2$ for all measurable $E \subset [0,1]$.
I have been able to show that $0 \le f < 1$ on $[0,1]$, but I cannot seem to get further than that.
On
There is an assumption missing: either you want $f\geq0$ or $\int_E|f|\leq m(E)^2$. Otherwise $f=-1$ satisfies the inequality.
Both assumptions I mention lead us to consider $f\geq0$. Then $\|f\|_1\leq1$. This allows us to apply Lebesgue Differentation: for each Lebesgue point of $f$, we have $$ 0\leq f(x)=\lim_{r\to 0}\frac1{m(B_r(x)}\int_{B_r(x)} f\leq\lim_{r\to0}\frac{m(B_r(x)^2}{m(B_r(x)}=0. $$
Martin Argerami's answer is correct, and in particular the problem should have assumed that $f\geq0$. But I think the use of Lebesgue's differentiation theorem is overkill. Here's a proof using only the fact that a non-negative measurable $f$ whose integral over $[0,1]$ is $0$ must be $0$ a.e. on $[0,1]$. (This fact is easily proved by considering the sets $A_n=\{x\in[0,1]:f(x)\geq\frac1n\}$ and showing that they have measure $0$ for all $n$.)
So I'll assume $f\geq0$ and also that $f$ is measurable (which is implicit in the problem since its integral is mentioned, but perhaps should have been explicit), and I'll prove that the integral of $f$ over $[0,1]$ is $0$. For any positive integer $n$, partition $[0,1]$ into $n$ intervals of length $\frac1n$. The integral of $f$ over any one of these intervals is, by hypothesis, at most $(\frac1n)^2$. So the integral of $f$ over the whole of $[0,1]$ is at most the sum of $(\frac1n)^2$ over all $n$ of the subintervals; that is, it is at most $(\frac1n)^2n=\frac1n$. Since this holds for all $n$ and since the integral of $f$ is nonnegative, we have $\int_{[0,1]}f=0$.