Let $E \subset [0,1]$ be a non-zero measurable set. Show that there exists an open interval $(a,b)\subset [0,1]$ such that $m(E\cap(a,b)) > \frac{1}{2}m(a,b) > 0$.
Proof: I will do it by contradiction and I want a proof verification.
Assume that
$m(E) > 0, \mbox{ and } \forall (a,b) \mbox{ open interval } m(E\cap (a,b)) \leq \frac{1}{2}m(a,b) \leq 0$
Consider $(a,b) = (0,1)$, then
$$ m(E\cap(0,1)) = m(E) > 0$$
That would contradict that $m(E) \leq 0$
I think that there is a mistake in my negation of the inequality because every measure has to be positive. How can I approach this ?
As the above comments point out, your proof is not correct. Here's an approach to the problem. Note that for any given $\epsilon>0$, there exists a finite family of disjoint intervals $I_j=(a_j,b_j)$ such that for $S=\cup_{j=1}^n I_j$, it holds $$ m(E\triangle S)<\epsilon. $$ This implies $$ m(E\cap S)\ge m(E)-m(E\triangle S)>m(E)-\epsilon. $$ On the other hand, $$\begin{eqnarray} m(E\cap S) = m\left(E\cap \left[\cup_{j=1}^n I_j\right]\right)&=&\sum_{j=1}^n m(E\cap I_j)\\&\le& \frac{1}{2}\sum_{j=1}^n m(I_j)=\frac{1}{2}m(S)\\&\le&\frac{1}{2}\left[m(E)+m(E\triangle S)\right]. \end{eqnarray}$$ Combining them, we have $$ m(E)-\epsilon\le\frac{1}{2}m(E)+\frac{\epsilon}{2}. $$ Since $\epsilon>0$ was arbitrary, we have $$ m(E)\le \frac{1}{2}m(E) $$ or $m(E)=0$. This contradicts $m(E)>0$.