I'm trying solving the following problem:
Let $f:[0,1]\to \Bbb{R}$ be a measurable question such that $f(x)>0$ a.e. Let $\{E_k\}_{k=1}^\infty\subset [0,1]$, a sequence of set such that $$\int_{E_k}f(x)\,dx\to0,\quad\mbox{when}\ k\to\infty.$$ Show that $m(E_k)\to0$, when $k\to\infty$.
I did a solution, but a realize I haven't used the fact that $E_k\subset[0,1]$, thus I'd like some hint or sketch for a solution.
Thanks!
My attempt of solution:
For each $m\in\Bbb{N}$ define $E^m_k=\{x\in E_k: f(x)>\frac{1}{m}\}$, since $f$ is measurable, $E_k^m$ is measurable too. Note that $E^m_k\subset E^{m+1}_k$, then $\lim m(E_k^m)=m(E_k)$. Given $\varepsilon>0$, there exists $N_1\in\Bbb{N}$ such that $$|m(E_k)-m(E_k^m)|<\frac{\varepsilon}{2},\ \forall m\geq N_1.$$ By other hand, $$m(E^m_k)=\int_{E^m_k}dx\leq m\int_{E^m_k}f(x)dx\leq m\int_{E_k}f(x)dx.$$ Since $\displaystyle\int_{E_k}f(x)dx\to0$ when $k\to\infty$, given $\varepsilon>0$, there exists $N_2\in\Bbb{N}$ such that $$\int_{E_k}f(x)dx<\frac{\varepsilon}{2m},\ \forall k\geq N_2.$$ Thus, $m(E_k^m)<\frac{\varepsilon}{2}, \ \forall k\geq N_2$.
Take $N=\max\{N_1,N_2\}$, then whenever $k,m\geq N$ we have $$|m(E_k)|<|m(E_k)-m(E_k^m)|+|m(E_k^m)|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$ We conclude that $m(E_k)\to0$ when $k\to\infty$.
(Observe that I haven't used that fact that $E_k\subset [0,1]$, therefore I think something is wrong!)
Your constant $N_1$ depends on $k$. Here is a simplified argument that corrects the error. The final inquality will be helpful.
Let $F_j = \{x : f(x) > 1/j\}$. Then $F_1 \subset F_2 \subset F_3 \subset \cdots$ and $m(\cup F_j) = 1$ since $f > 0$ almost everywhere. Apply continuity from below to see $m(F_j) \to 1$.
Fix $\epsilon > 0$. Since $m(F_j) \to 1$, there exists an index $j$ with the property that $m([0,1] \setminus F_j) < \epsilon$. For any measurable set $E$, $$ m(E) = m(E \cap F_j) + m(E \setminus F_j) \le m(E \cap F_j) + \epsilon.$$ Since $$m(E \cap F_j) = \int_{E \cap F_j} \, dx \le j \int_{E \cap F_j} f \, dx \le j \int_E f \, dx$$ you have in particular $$ m(E) \le j \int_E f \, dx + \epsilon$$ for any measurable $E \subset [0,1]$.