Prove that $\inf\left\{\int_E f d\mu : E \in \mathcal{M}, \mu(E)\geq 0 \right\}>0$ 
Problem. Suppose that $(X,\mathcal{M},\mu)$ is a nonzero finite measure space and that $f$ is a positive measurable function on $X$. Let $\alpha\in (0,\infty)$ with $0<\alpha<\mu(X)<\infty$ be given. Then prove that $$\inf\left\{\int_E f d\mu : E \in \mathcal{M}, \mu(E)\geq \alpha  \right\}>0$$  

I have attempted this problem and have not been able to solve. In my imagination, it can be $0$ since it seems there is no restriction for set to take any value on $(0,\infty)$ and $\inf$ is limit point of it. Actually, this is my first time to face this kind of problem. I have no idea. I hope to get some help to solve this problem. Thank you in advance.
 A: For each positive integer $n$, let $U_n$ denote the preimage of $\left(\dfrac1n,\infty\right)$ under $f$.  We know that $$\bigcup_{n=1}^\infty\,U_n=X$$ since $f$ is a positive function.  Thus, there exists $N\in\mathbb{Z}_{>0}$ for which
$$\mu(U_n)\geq\mu(X)-\frac{\alpha}{2}$$
for all integers $n\geq N$.
Let $E$ be a measurable subset of $X$ with $\mu(E)\geq \alpha$.  We see that
$$\begin{align}\mu(E\cap U_n)&=\mu(E)+\mu(U_n)-\mu(E\cup U_n)\geq \mu(E)+\mu(U_n)-\mu(X)\\
&\geq \alpha +\left(\mu(X)-\frac{\alpha}{2}\right)-\mu(X)= \frac{\alpha}{2}\end{align}$$
for all integers $n\geq N$.  Thus,
$$\int_{E}\,f\,\text{d}\mu\geq \int_{E\cap U_N}\,f\,\text{d}\mu\geq \frac{\alpha}{2N}>0\,.$$
That is,
$$\inf\left\{\int_E\,f \,\text{d}\mu \,\Big|\, E \in \mathcal{M}\text{ and } \mu(E)\geq \alpha  \right\}\geq \frac{\alpha}{2N}>0\,.$$
P.S.  The assumption that $\mu(X)<\infty$ is quite important.  Take $X:=\mathbb{R}_{\geq 0}$, $\mathcal{M}$ the collection of Borel-measurable sets of $X$, $\mu$ the usual Lebesgue measure on $X$, and $f:X\to\mathbb{R}_{>0}$ defined by $$f(x):=\exp(-x)\text{ for all }x\geq 0\,.$$
Then, for any $\alpha\in[0,\infty]$,
$$\inf\left\{\int_E\,f \,\text{d}\mu \,\Big|\, E \in \mathcal{M}\text{ and } \mu(E)\geq \alpha  \right\}=0\,.$$
This is because
$$\lim_{n\to\infty}\,\int_n^\infty\,\exp(-x)\,\text{d}x=\lim_{n\to\infty}\,\exp(-n)=0\,.$$
