f is L(A) implies finite almost everywhere?

**younhock** · November 22nd 2011, 08:13 AM

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

Please Help ~~

**girdav** · November 22nd 2011, 08:31 AM

$\left\{f=\pm\infty\right\}=\bigcap_{n\geq 1}\left\{|f|\geq n\right\}$ and $m(\left\{|f|\geq n\right\})\leq \frac 1n\int_A|f|dm$ . Since $m(A)<\infty$ , $m(\left\{f=\pm\infty\right\})=\lim_{n\to\infty}m( \left\{|f|\geq n\right\})$ and you can conclude.

**younhock** · November 22nd 2011, 08:44 AM

Originally Posted by girdav

$\left\{f=\pm\infty\right\}=\bigcap_{n\geq 1}\left\{|f|\geq n\right\}$ and $m(\left\{|f|\geq n\right\})\leq \frac 1n\int_A|f|dm$ . Since $m(A)<\infty$ , $m(\left\{f=\pm\infty\right\})=\lim_{n\to\infty}m( \left\{|f|\geq n\right\})$ and you can conclude.

$\bigcap_{n\geq 1}\left\{|f|\geq n\right\}$ ?? or union??

**girdav** · November 22nd 2011, 08:49 AM

Originally Posted by younhock

$\bigcap_{n\geq 1}\left\{|f|\geq n\right\}$ ?? or union??

No, it's the intersection, since if $|f(x)|=+\infty$ then $|f(x)|$ in greater than each integer.

**younhock** · November 22nd 2011, 08:51 AM

Originally Posted by girdav

No, it's the intersection, since if $|f(x)|=+\infty$ then $|f(x)|$ in greater than each integer.

Oh yes i get this. But why is $m(\left\{|f|\geq n\right\})\leq \frac 1n\int_A|f|dm$ ?

**girdav** · November 22nd 2011, 08:53 AM

$n\cdot m(\left\{|f|\geq n\right\})\leq \int_{\left\{|f|\geq n\right\}}|f|dm\leq \int_{A}|f|dm$

**younhock** · November 22nd 2011, 08:54 AM

Originally Posted by girdav

$n\cdot m(\left\{|f|\geq n\right\})\leq \int_{\left\{|f|\geq n\right\}}|f|dm\leq \int_{A}|f|dm$

OKAY!! THANKS a lot!!!!!

**girdav** · November 22nd 2011, 09:41 AM

And note that the result is true for a $\sigma$ -finite measured space $(X,\mathcal A,\mu)$ , i.e. a space such that we can find a countable partition of $X$ into sets of finite measure (for example $\mathbb R$ with Lebesgue measure).

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