# Integration of nonnegative funtion, Folland Real Analysis

Suppose $f$ is a nonnegative measurable function on a measure space $(X,M,\mu)$ satisfying $\int f d\mu < \infty$. Show that for every $\epsilon > 0$ there exists a $\delta > 0$ such that if $\mu(E) < \delta$ then $$\int_{E}f d\mu < \epsilon$$

Attempted proof: Let $E_n = \{f \geq 1/n\}$, then if we apply Fatou's lemma to $f_n:= f 1_{E_n}$, we have $$\lim_{n\rightarrow \infty}\int_{E_n}f d\mu \leq \int_{E_n} \lim_{n\rightarrow \infty}f d\mu \leq \int_{E_n}\lim_{n\rightarrow \infty} \inf f d\mu$$ $$\leq \lim_{n\rightarrow \infty}\inf\int_{E_n} f d\mu = \lim_{n\rightarrow\infty}\inf\int f 1_{E_n} d\mu$$

I am not sure if this is the correct approach or if I should be using a different theorem, any suggestions is greatly appreciated.

Perhaps a better approach: Let $\phi$ be a simple function in $L^{+}$ with standard representation $\phi = \sum_{1}^{N} a_j 1_{E_n}$, let $E_n = \{f \geq 1/n\}$ then $$\int \phi d\mu = \sum_{1}^{N}a_j \mu(E_n)$$ as $n\rightarrow \infty$ we have $$\int \phi d\mu = 0$$ then we can select an $n_0$ such that $\int_{E_{n_0}} < \epsilon/2$ and proceed....

A little stuck on this problem as you can see

• Now thinking about it more, I think I should use the Dominated convergence theorem, although I think Fatou's lemma or the monotone convergence theorem would work as well. Commented Dec 1, 2015 at 2:53
• Try proving a simple case first: suppose that, in addition to the assumptions here, $f$ is also bounded. That shouldn't cause too much grief. Then think about where to go next. Commented Dec 1, 2015 at 4:16

Let $B_n=\{x:x\in X,\:n-1<f(x)\leqslant n\}$ and $C=\{x:x\in X,\:f(x)>n\}$. We have $$\int_Xfd\mu=\int_{\bigcup_{n=1}^{\infty}B_n}fd\mu=\sum_{n=1}^{\infty}\int_{B_n}fd\mu$$ Since $\sum_{n=1}^{\infty}\int_{B_n}fd\mu$ is absolute convergent, given $\epsilon>0$ there is a $N$ such that for any $n>N$ $$\sum_{n=N+1}^{\infty}\int_{B_n}fd\mu=\int_{\bigcup_{n=N+1}^{\infty}B_n}fd\mu<\epsilon/2\tag{1}$$ Let $$B=\bigcup_{n=1}^{N}B_n\quad\text{and}\quad C=\bigcup_{n=N+1}^{\infty}B_n=X-B$$ Take $\delta=\dfrac{\epsilon}{2(N+1)}$. Then for any $E\subset X$ such that $\mu(E)<\delta$, by $(1)$ $$\int_Efd\mu=\int_{E\cap X}fd\mu=\int_{E\cap (B\cup C)}fd\mu=\int_{E\cap B}fd\mu+\int_{E\cap C}fd\mu\leqslant N\int_Ed\mu+\int_Cfd\mu<N\dfrac{\epsilon}{2(N+1)}+\dfrac{\epsilon}{2}<\epsilon$$
• I am not sure if your approaches are completely wrong. But in order to prove this one, you need to prove that $\mu(\{x:x\in X,\:f(x)>n\})$ is arbitrary small because the integral is bounded for $\{x:x\in X,\:f(x)<n\}$. So $\mu(\{x:x\in X,\:f(x)>1/n\})$ won't help. Commented Dec 1, 2015 at 23:00
• are you missing the $f$, in the $\leq N\int_{E} d\mu$ part? Commented Dec 6, 2015 at 19:55
• No. $\int_{E\cap B}fd\mu\leqslant N\int_Ed\mu$ since $f\leqslant N$ for $x\in B$. Commented Dec 7, 2015 at 0:06
Here's another approach, by contradiction. Assume then that for some $\epsilon_0>0$, and for any $\delta >0$, there exists $E_\delta$ with $m(E_\delta) <\delta$, yet $\int_{E_{\delta}} f > \epsilon_0$. This gives a sequence $E_{1/n}$ with $m(E_{1/n})< 1/n \to 0$, such that $\int_{E_{1/n}} f \ge \epsilon_0$. Then ${\bf 1}_{E_{1/n}} f\to 0$ in measure, and dominated by $|f|$. By dominated convergence theorem (for convergence in measure), we have $\int_{E_{1/n}} f \to 0$, a contradiction.