# Integration on measure theory

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.

1. 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).
2. 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 $$\mu\{x\mid f(x)=+\infty\}=\lim_{n\to \infty}\mu\{x\mid f(x)\geq n\}=0.$$
Thank you for your help. <br> But I'm not sure why these two note: $$\lambda\{x\mid f(x)\geq \frac 1n\}\leq n\int fd\lambda=0,$$ $$\lambda\{x\mid f(x)\geq n\}\leq \frac 1n\int fd\lambda,$$ –  japee May 8 '12 at 11:53
If $A$ is a real number, $A\lambda(x\mid f(x)\geq A)\leq \int f(x) \lambda$. –  Davide Giraudo May 8 '12 at 12:25
Take $f(x)=\frac 1x$ on $(0,1)$. –  Davide Giraudo Oct 1 '13 at 8:26