LET f is Lebesgue Integrable on A where m(A) is finite, Prove that f must be finite almost everywhere on A.

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- Nov 22nd 2011, 08:13 AMyounhockf is L(A) implies finite almost everywhere?
LET f is Lebesgue Integrable on A where m(A) is finite, Prove that f must be finite almost everywhere on A.

Please Help ~~(Crying) - Nov 22nd 2011, 08:31 AMgirdavRe: f is L(A) implies finite almost everywhere?
$\displaystyle \left\{f=\pm\infty\right\}=\bigcap_{n\geq 1}\left\{|f|\geq n\right\}$ and $\displaystyle m(\left\{|f|\geq n\right\})\leq \frac 1n\int_A|f|dm$. Since $\displaystyle m(A)<\infty$, $\displaystyle m(\left\{f=\pm\infty\right\})=\lim_{n\to\infty}m( \left\{|f|\geq n\right\})$ and you can conclude.

- Nov 22nd 2011, 08:44 AMyounhockRe: f is L(A) implies finite almost everywhere?
- Nov 22nd 2011, 08:49 AMgirdavRe: f is L(A) implies finite almost everywhere?
- Nov 22nd 2011, 08:51 AMyounhockRe: f is L(A) implies finite almost everywhere?
- Nov 22nd 2011, 08:53 AMgirdavRe: f is L(A) implies finite almost everywhere?
$\displaystyle n\cdot m(\left\{|f|\geq n\right\})\leq \int_{\left\{|f|\geq n\right\}}|f|dm\leq \int_{A}|f|dm$

- Nov 22nd 2011, 08:54 AMyounhockRe: f is L(A) implies finite almost everywhere?
- Nov 22nd 2011, 09:41 AMgirdavRe: f is L(A) implies finite almost everywhere?
And note that the result is true for a $\displaystyle \sigma$-finite measured space $\displaystyle (X,\mathcal A,\mu)$, i.e. a space such that we can find a countable partition of $\displaystyle X$ into sets of finite measure (for example $\displaystyle \mathbb R$ with Lebesgue measure).