let $f\ge 0$ be a measurable function s.t. $\int_R fdm=\infty$, show that for any M>0 there is a real measurable function g, and $0\le g \le f$ and the following hold: $\int_R g dm \ge M$ and g is bounded and m(x:g(x) $\ne$ 0)< $\infty$.
I think g is just a simple function s, such that 2M > $\int_R s dm$ > M. s is bounded since s(x)=$\sum_{i=1}^n c_iI_{E_i}$ <$\sum_{i=1}^n ci$.
Would this be correct?
Yes, the integral is the limit of the integrals of simple functions smaller than $f$. Therefore, if the integral in infinite, there should be arbitrary large integrals of simple functions smaller than $f$. Take one that is larger than $M$.