Thread: 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?

$\displaystyle \left\{f=\pm\infty\right\}=\bigcap_{n\geq 1}\left\{|f|\geq n\right\}$ 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.

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

Originally Posted by girdav
$\displaystyle \left\{f=\pm\infty\right\}=\bigcap_{n\geq 1}\left\{|f|\geq n\right\}$ 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.
$\displaystyle \bigcap_{n\geq 1}\left\{|f|\geq n\right\}$ ?? or union??

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

Originally Posted by younhock
$\displaystyle \bigcap_{n\geq 1}\left\{|f|\geq n\right\}$ ?? or union??
No, it's the intersection, since if $\displaystyle |f(x)|=+\infty$ then $\displaystyle |f(x)|$ 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 $\displaystyle |f(x)|=+\infty$ then $\displaystyle |f(x)|$ in greater than each integer.
Oh yes i get this. But why is $\displaystyle m(\left\{|f|\geq n\right\})\leq \frac 1n\int_A|f|dm$ ?

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

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

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

Originally Posted by girdav
$\displaystyle n\cdot m(\left\{|f|\geq n\right\})\leq \int_{\left\{|f|\geq n\right\}}|f|dm\leq \int_{A}|f|dm$
OKAY!! THANKS a lot!!!!!

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

And note that the result is true for a $\displaystyle \sigma$-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).