# f is L(A) implies finite almost everywhere?

• Nov 22nd 2011, 09:13 AM
younhock
f 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.

• Nov 22nd 2011, 09:31 AM
girdav
Re: f is L(A) implies finite almost everywhere?
$\left\{f=\pm\infty\right\}=\bigcap_{n\geq 1}\left\{|f|\geq n\right\}$ 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.
• Nov 22nd 2011, 09:44 AM
younhock
Re: f is L(A) implies finite almost everywhere?
Quote:

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??
• Nov 22nd 2011, 09:49 AM
girdav
Re: f is L(A) implies finite almost everywhere?
Quote:

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.
• Nov 22nd 2011, 09:51 AM
younhock
Re: f is L(A) implies finite almost everywhere?
Quote:

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 $m(\left\{|f|\geq n\right\})\leq \frac 1n\int_A|f|dm$ ?
• Nov 22nd 2011, 09:53 AM
girdav
Re: f is L(A) implies finite almost everywhere?
$n\cdot m(\left\{|f|\geq n\right\})\leq \int_{\left\{|f|\geq n\right\}}|f|dm\leq \int_{A}|f|dm$
• Nov 22nd 2011, 09:54 AM
younhock
Re: f is L(A) implies finite almost everywhere?
Quote:

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!!!!!
• Nov 22nd 2011, 10:41 AM
girdav
Re: f is L(A) implies finite almost everywhere?
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).