A nice property of $L_1$ function in $[0,1]$.

Question

Let $f$ is in $L^1([0,1],m)$, $m$ is a Lebesgue measure and suppose that $f(x)>0$ for all $x$. Show that for any $ 0<\epsilon<1 $ there exists $\delta$ >0 so that $\int_{E} f(x)dx\ge \delta$ for any set $E\subset[0,1]$ with $m(E)=\epsilon$

I have done a similar problem something like if $f$ is in $L^1$ then we have $\epsilon$ and $\delta$ definition. However, this problem seems a bit different even kinda of opposite in the sense of $\epsilon$ and $\delta$. I could get my head around it.

Answer 1

Hint: take $\eta > 0$ so that $m\left(\{x: 0 < f(x) < \eta \}\right) \le \epsilon/2$.

Answer 2

For each integer $n$ let $$A_n= f^{-1} [2^{n-1},2^n)=\{x\in [0,1]:f(x)\in [2^{n-1},2^n)\}.$$ Let $B_n=m(A_n)$ where $m$ is Lebesgue measure.$$\text { We have }\lim_{k\to \infty}\sum_{n=-k}^{n=k}B_n= \sum_{n\in Z}B_n=1.$$ Now given $\epsilon \in (0,1)$, choose $k\in Z^+$ for which $$\sum_{n=-k}^{n=k}B_n>1-\epsilon /2.$$ $$\text {Now let } S=\cup_{|n|\leq k}A_n$$ $$\text{and let } \delta=(\epsilon /2)2^{-k}.$$ If now $E$ is a measurable subset of $[0,1]$ and $m(E)\geq \epsilon$ then $m(E\cap S)>\epsilon /2$ so we have $$\int_Ef(x) dx\geq \int_{E\cap S}f(x)dx\geq m(E\cap S)2^{-k}>(\epsilon /2)2^{-k}=\delta.$$

Answer 3

Note that all the conditions in Lebesgue integral of non-negative are satisfied.

You may refer to my solution.