Problem:
Let $f>0$, $f$ is Lebesgue measurable on $[0,1]$. Prove: $$\text{For all } \alpha\in(0,1), \inf\left\{\int_E f\,\mathrm dm: \text{ Measurable set } E\subset [0,1] \text{ and } m(E)>\alpha\right\}>0,$$
where $m$ denotes the Lebesgue measure.
I want to prove this by showing $\int_e fdm$ is bigger than a positive number that has nothing to do with $e$. Here is my attempt:
Assume $F_n=\{x\in [0,1]: f(x)>1/n\}$, then $F_n$ increases with regard to $n$, and $\lim_{n\to\infty}m(F_n)=m(\lim_{n\to\infty}F_n)=1$. Then $\forall \alpha\in(0,1), \exists n_0\ s.t. 1-m(F_{n_0})<\alpha$. Now $\int_efdm\geq\int_{e\cap F_{n_0}}fdm>1/n_0*m(e\cap F_{n_0})$. Then it suffices to show $m(e\cap F_{n_0})>0$. Since we have $m(e)>\alpha$ and $m(F_{n_0})>1-\alpha$, can we prove $m(e\cap F_{n_0})>0$ ?
Best Answer
The bound $m(F_{n_0})>1-\alpha$ is not enough for an explicit bound. Instead, consider $$\exists n_0: \frac{\alpha}{2}\ge m(F_{n_0}^c)> 0\\ m(e\cap F_{n_0})=m(e)-m(F_{n_0}^c\cap e)\ge m(e)-\frac{\alpha}{2}\ge\frac\alpha2$$ Where $F_{n_0}^c$ is the complement of $F_{n_0}$ in $[0,1]$. The result follows