Lebesgue integral of non-negative

Question

Assume that $f: [0,1] \rightarrow [0,\infty)$ is a Lebesgue measureable function such that $f(x) > 0$ for a.e $x.$. Show that for every $\epsilon >0$ there is $\delta >0$ such that for every Lebesgue measurable set $E$ with Lebesgue measure $\lambda(E) \geq \epsilon$ we have $\int_{E} f d \lambda \geq \delta$.

An idea: We proceed by contradiction: Suppose there exists an $\epsilon_{0}$ and a sequence of measurable sets $\{E_{n} \}_{n \in \mathbb{N}}$, such that for all n, we have $\lambda (E_{n}) \geq \epsilon_{0}$ while $2^{-n} > \int_{E_{n}}f d\lambda$. For each $n$, define $F_{n} = \cup_{k=n}^{\infty}E_{k}$, which is a decreasing sequence of sets. $F = \cap_{n=1}^{\infty}F_{n} = \lim_{n \to \infty} \sup E_{n}$.

By the continuity of $\lambda$ from above, we have $\lambda (F) \geq \lim_{n \to \infty} \sup \lambda(E_{n}) \geq \epsilon_{0}$.

I do not see how this leads to a contradiction. Perhaps there is a better way. I will appreciate a hint or a solution.

Answer 1

Let $1>\epsilon>0$ and $F_R = \{x\in [0,1] : 0 < f(x) \le R\}$. Let $c >0$ be given by $$c = \inf \{ R>0: \lambda (F_R) \ge \epsilon\}.$$ As $f(x) >0$ a.e., we have $c >0$. $$\bigcap_{R \ge c} F_R = F_c,$$ we have $\lambda (F_c) \ge \epsilon$ and $\lambda(F_d) <\epsilon$ for all $d <c$. Write $$F_c = \{f(x) = c\} \cup \{f(x) < c\}.$$ Note that $\lambda(\{f(x) < c\}) \le \epsilon$. Let $D$ be such that $$ \{ f(x) < c\} \subset D \subset \{ f(x) \le c\} = F_c\ \ \ \text{and } \lambda (D) = \epsilon.$$ Let $\delta = \int_{D} f d\lambda >0$. Let $E$ be any measurable set in $[0,1]$ and $\lambda (E)\ge \epsilon$. Then

$$\begin{split} \int_E f d\lambda &= \int_{E\cap D} f d\lambda + \int_{E\setminus D} fd\lambda \\ &\ge \int_{E\cap D} f d\lambda + \int_{E\setminus D} c d\lambda \\ &= \int_{E\cap D} f d\lambda + \lambda(E\setminus D)c \end{split}$$ As $$\begin{split}\lambda (E\setminus D) + \lambda (E\cap D) &= \lambda (E) \\ &\ge \epsilon \\ &= \lambda (D) \\ &= \lambda (D\cap E) +\lambda (D\setminus E)\\ \Rightarrow \lambda (E\setminus D) &\ge \lambda (D\setminus E) . \end{split}$$ So $$\begin{split} \int_E fd\lambda &\ge \int_{E\cap D} f d\lambda + \lambda(D\setminus E)c\\ &= \int_{E\cap D} f d\lambda + \int_{D\setminus E} cd\lambda \\ &\ge \int_{E\cap D} f d\lambda + \int_{D\setminus E} fd\lambda \\ &= \int_D fd\lambda = \delta . \end{split}$$

Answer 2

For each $n\in\mathbb{N}$, let $A_{n}=\{x\in[0,1]\mid f(x)\geq\frac{1}{n}\}$. Then $A_{1}\subseteq A_{2}\subseteq\ldots$ and $\cup_{n}A_{n}=\{x\mid f(x)>0\}$ which has measure 1. Denote the Lebesgue measure on $[0,1]$ by $\lambda$. We have $1=\lambda(\cup_{n}A_{n})=\lim_{n}\lambda(A_{n})$.

Let $\varepsilon>0$ be given. Choose $n$ such that $\lambda(A_{n}^{c})<\frac{\varepsilon}{2}$. (Here, the complement is relative to the set $[0,1]$.) Let $\delta=\frac{\varepsilon}{2}\cdot\frac{1}{n}$. Let $E\subseteq[0,1]$ be a measurable set with $\lambda(E)\geq\varepsilon$. Observe that $$ \lambda(E\cap A_{n})=\lambda(E)-\lambda(E\cap A_{n}^{c})\geq\varepsilon-\frac{\varepsilon}{2}=\frac{\varepsilon}{2} $$ and $f\geq\frac{1}{n}$ on $A_{n}\cap E$. Therefore, $$ \int_{E}f\,d\lambda\geq\int_{E\cap A_{n}}f\,d\lambda\geq\frac{1}{n}\lambda(E\cap A_{n})\geq\frac{1}{n}\cdot\frac{\varepsilon}{2}=\delta. $$