Real Valued Lebesgue Intergal

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!

Re: Real Valued Lebesgue Intergal

There's an issue with the way you've set up the argument. You're not assuming that f is a nonnegative function, so it does not follow that the integral over the whole space is necessarily larger than the integral over a measurable subset. This isn't a big issue, but it does require some comment. Otherwise, it looks like you're on the right track. Just remember that in this context, infinity is an extended real number which has a defined set of algebraic rules and is not a limit.