Problem:
$f$ is integrable on $[0,2]$, for any measurable set $E \subseteq [0,2]$, when $m(E) = 1$, $\int_E fdm = 0$ prove $f = 0$ $a.e.$ on $[0,2]$
I think the following theorem might be of help, but I'm not sure.
Theorem. If $f$ is integrable on $[a,b]$ and $\int_a^x f(t)dt = 0$ $\forall x \in [a,b]$, then $f = 0$ $a.e.$ on $[a,b]$.
The proof is in this link.
Let $E_- = \{f < 0\}$ and $E_+ = \{f > 0\}$ and $A = [0,2] \setminus E_- \setminus E_+ = \{f = 0\}$. Suppose that $m(E_-) \geq 1$ or $m(E_+) \geq 1$. In either case we get a contradiction when integrating over the larger set. Thus $m(E_-),m(E_+) < 1$. Now we integrate over $E_- \cup A$ and $E_+ \cup A$ to find that in fact $m(E_-) = m(E_+) = 0$.