The next problem appears in the book "Real analysis" by Thomson/Bruckner in the chapter of Integrable functions. My proof seems good to me but I'm not confident enough with my proofs in this subject. Here, $(X, \sigma, \mu)$ is a measurable space. I use the following:
Definition Given a succession $(E_n)$ of measurable sets, $\limsup E_n = \{x \in X : x$ belongs to $E_j$ for infinitely many $j \}$
Theorem: if $(E_n)$ is a succession of measurable sets and $\mu (\cup E_n) < \infty$ then $\limsup \mu(E_n) \leq \mu(\limsup E_n)$
Problem: Suppose that $f \in L_1(X)$, that $f(x)>0$ for all $x \in X$, and that $0 < \alpha < \mu(X) < \infty $. Prove that $$ \inf \left\{ \int_E f d\mu : \mu (E) \geq \alpha\right\} > 0. $$
My attempt:
Suppose that $\inf \left\{ \int_E f d\mu : \mu (E) \geq \alpha\right\} = 0$. Then, we can choose a succession $(E_n)_{n \geq 1}$ such that $\mu(E_n) \geq \alpha$ and $$ \int_{En} f < \frac{\alpha}{2^n}.$$ Then, $f|_{E_n} < \frac{1}{2^n} $. For if $f|_{E_n} \geq \frac{1}{2^n} $ then we have $$ \frac{\alpha}{2^n} > \int_{E_n} f d \mu \geq \int_{E_n} \frac{d \mu}{2^n} = \frac{\mu(E_n)}{2^n} \geq \frac{\alpha}{2^n},$$ a contradiction.
Now, if there exist $ x \in X$ such that $x \in \limsup E_n$ then $f(x)<\frac{\alpha}{2^j}$, for infinitely many $j's$. So we must have $f(x) = 0$, contradicting that $f > 0$. Therefore $\limsup E_n = \emptyset$ and $\mu (\limsup E_n)=0$. However $\limsup \mu(E_n) \geq \alpha$ because $\mu(E_n) \geq \alpha$ for every $n$. And because we are in a space of finite measure, we can aply the above theorem and obtain $$ 0 < \alpha \leq \limsup \mu(E_n) \leq \mu (\limsup E_n) = 0. $$ So $\alpha = 0$ is our final contradiction.
For $n\in\mathbb{N}$, define $A_n=\{x:f(x)>\frac{1}{n}\}$ note that $A_n\uparrow X$ (because $f(x)>0$), so $\mu(A_n)\uparrow\mu(X)$. Hence, there is some $m$ big enough such that $\mu(X)-\mu(A_m)=\mu(X-A_m)<\alpha/2$.
Therefore, for any set $E$ with $\mu(E)>\alpha$, you will have that $\mu(E\cap A_m)\geq \mu(E)-\mu(X-A_m)>\frac{\alpha}{2}$.
Hence, $\int_Ef d\mu\geq \int_{E\cap A_m}f d\mu\geq\int_{E\cap A_m}\frac{1}{m}d\mu=\frac{1}{m}\mu(E\cap A_m)>\frac{\alpha}{2m}$.
So the infimum is at least $\frac{\alpha}{2m}>0$.