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

1. ## 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.

2. ## 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.

3. ## Re: f is L(A) implies finite almost everywhere?

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??

4. ## Re: f is L(A) implies finite almost everywhere?

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.

5. ## Re: f is L(A) implies finite almost everywhere?

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 $m(\left\{|f|\geq n\right\})\leq \frac 1n\int_A|f|dm$ ?

6. ## Re: f is L(A) implies finite almost everywhere?

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

7. ## Re: f is L(A) implies finite almost everywhere?

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!!!!!

8. ## 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).