Hello!

I'm trying to show that if f is a measurable function on some space X, and if

$\displaystyle \int{fd\mu} < \infty $

then $\displaystyle \mu\{x \in X : f(x) = \infty\} = 0 $

In other words, trying to show f is real-valued (not infinity) mu-almost everywhere.

I thought maybe the contrapostive would work, ie,

Presume $\displaystyle \mu\{x \in X : f(x) = \infty\} > 0 $.

Call the set $\displaystyle \{x \in X : f(x) = \infty\} = B$ (for "big" valued, hehe)

Then $\displaystyle \int_B{fd\mu} = \int{f\chi_B d \mu}$

If I can just show $\displaystyle \int{f\chi_B d \mu} = \infty $, then since $\displaystyle B \subset X$,

$\displaystyle \int_X{f d \mu} = \int{f d \mu} = \infty $ as well.

So I am just wondering if the above looks alright, and what's left is how to possibly show that $\displaystyle \int{f\chi_B d \mu} = \infty $ for $\displaystyle \{x \in X : f(x) = \infty\} = B$ and $\displaystyle \mu(B) > 0 $?!

Thanks very much!