A past qual question from my university reads:
Let $f$ be an integrable function satisfying $\int_0^1 f(x)dx=0$. Prove that there are intervals $I$ of arbitrarily small positive length such that
$$\int_I f(x)dx=0$$
I'm not sure how to approach the problem. One has that $\nu(E)=\int_E f(x)dx$ is a signed measure with a Hahn decomposition of $[0,1]=P\cup N$ where $f\geq 0$ on P and $f\leq 0$ on N. But I can't seem to be able to come up with a way of finding an interval with the desired property.
Best Answer
We can show that for every $n \ge 2$ there is an interval $I_n$ of length between $1/n$ and $2/n$ s.t. $\int_{I_n} f(x)dx=0$
Fix $n \ge 2$ and consider $a_k=\int_{\frac{k}{n}}^{\frac{k+1}{n}}f(x)dx, k=0,1..,n-1$; since $\sum a_k=0$ we either have some $a_k =0$ so done or there are consecutive $a_ka_{k+1} <0$ for some $k \le n-2$; wlog assume $a_k >0, a_{k+1} <0, a_k+a_{k+1} >0$ since if $a_k+a_{k+1}=0$ we are again done, while the other cases are treated as below with the obvious changes.
Then $g(a)=\int_a^{\frac{k+2}{n}}f(x)dx$ is a continuos function for $\frac{k}{n} \le a \le \frac{k+1}{n}$ and $g(\frac{k}{n}) >0, g(\frac{k+1}{n}) <0$ so there is an $a_n, \frac{k}{n} \le a_n \le \frac{k+1}{n}, g(a_n)=0$;
letting $I_n=[a_n,\frac{k+2}{n}]$ we are done since $\int_{I_n} f(x)dx=0, 1/n < |I_n| < 2/n$