Hey guys thank you for helping me, It's vanishing property and finiteness property. and I wanna prove:
$f\ge0$ and $f$ is measurable. Then,
1. $\int fd\lambda=0 \Longleftrightarrow \{x|f(x)>0\}$ is a null set.
2. $\int fd\lambda<\infty \Longrightarrow \{x|f(x)=\infty\}$ is a null set.
I don't know how can I approach. please help me. thx
Integration on measure theory
1
$\begingroup$
real-analysis
measure-theory
-
3Is this homework? – 2012-05-06
1 Answers
2
- If $\{x\mid f(x)>0\}$ is a null set, then $f=0$ almost everywhere and so its integral is $0$. Conversely, assume that $\int fd\lambda=0$. Then writing $\{x\mid f(x)>0\}=\bigcup_{n\geq 1}\{x\mid f(x)\geq \frac 1n\}$ and noticing that $\lambda\{x\mid f(x)\geq \frac 1n\}\leq n\int fd\lambda=0,$ we have written $\{x\mid f(x)> 0\}$ as a countable union of sets of measure $0$, so this set has measure $0$ (an is measurable).
- We can write $\{x\mid f(x)=+\infty\}=\bigcap_{n\geq 1}\{x\mid f(x)\geq n\}$. Since $\mu\{x\mid f(x)\geq n\}\leq \frac 1n\int fd\lambda,$ we have, because all the involved sets have finite measure $\mu\{x\mid f(x)=+\infty\}=\lim_{n\to \infty}\mu\{x\mid f(x)\geq n\}=0.$
-
0Take $f(x)=\frac 1x$ on $(0,1)$. – 2013-10-01