Assume that $f : [0, 1] \to (0,\infty)$ is a Lebesgue measurable function. Show that for every $\epsilon > 0$ there is $\delta > 0$ such that for every Lebesgue measurable set $E$ with Lebesgue measure $λ(E) ≥ \epsilon$ we have $$\int_Efd\lambda\geq \delta$$
I tried to do it, but unable to finish the last part . (I tried to write down all the steps to see if there is any part wrong)
My attempt:
let $E = \bigcup_{k=1}^\infty E_k$ where $E_k $ is disjoint subset of $E$. Since $f$ is measurable there exists a non-decreasing sequence of simple function $S_n = \sum_{k=1}^n \alpha_k \chi_{E_k}$ with $\alpha_k \geq$ converging pointwise to $f$. So using MCT we can write;
$$\int_Efd\lambda = \int_E \lim_{n\to \infty} S_n d\lambda = \lim_{n\to \infty}\int_E S_n d\lambda = \lim_{n\to \infty}\int_{[0,1]} \sum_{k=1}^n \alpha_k \chi_{E_k} d\lambda $$ $$ = \lim_{n\to \infty} \sum_{k=1}^n \alpha_k \int_0^1\chi_{E_k} d\lambda = \sum_{k=1}^\infty \alpha_k \lambda (E_k)$$
can I conclude the result using below? (that there exist $\alpha\geq 0 $) such that $$\sum_{k=1}^\infty \alpha_k \lambda (E_k) \geq \sum_{k=1}^\infty \alpha \lambda (E_k) = \alpha \sum_{k=1}^\infty \lambda (E_k) = \alpha \lambda(E) \geq \alpha \epsilon = \delta $$
$\textbf{Edit}:$ Is this proof also would work?
suppose by contradiction that $\int_Efd\lambda=0$ , given that $λ(E) ≥ \epsilon$ we would have that $f=0$ a.e. where contradicts the assumpion $f : [0, 1] \to (0,\infty)$ .
You may try to prove by contradiction. Here is a straightforward approach I have in mind. Suppose the conclusion is false. Namely, there exists $\epsilon>0$ such that $\forall\ \delta>0$, we can find a set $E_\delta\subseteq [0,1]$ with $\lambda(E_\delta)\geq\epsilon$ satisfying $$ \int_{E_\delta} f d\lambda<\delta. $$ We may choose $\delta = 1/2^n$, for $n = 1, 2, ...$. Then we have construted a sequence $\{E_n\}_{n = 1}^\infty$ of sets with $\lambda(E_n)\geq\epsilon$, for all $n$, but $$ \int_{E_n} f d\lambda<1/2^n. $$ Now consider $E = \limsup E_n:=\bigcap_{k = 1}^\infty \bigcup_{n = k}^\infty E_n$. Specifically, consider $f_k = f\cdot\chi_{\bigcup_{n = k}^\infty E_n}$, $k = 1, 2, ...$, then $$ \int_{[0,1]} f_k d\lambda\leq \sum_{n = k}^\infty 1/2^n<\infty. $$ Therefore, $|f_k|=f_k\leq f_1=|f_1|$ with $f_1$ Lebesgue integrable and $f_k\to f\cdot \chi_E$ pointwisely. By dominated convergence theorem, $$ \int_E f = \int \lim_{k\to\infty} f_k = \lim_{k\to\infty}\int f_k=0. $$ But $\lambda(E)\geq\epsilon$ and $f$ is nonnegative. This means $f = 0$ a.e. on $E$, contradicting to $f>0$ on $[0,1]$.