Prove that $\int g\ {\rm d} \mu<\infty$?

Question

Let $(S,\mathcal{A},\mu)$ be aprobability space and $g\ge 0$ with $\int g\ln^+\ln^+ g {\rm d} \mu<\infty$. Can you help me prove that $\int g\ {\rm d} \mu<\infty$?

Answer 1

First, observe that, for every $M'\gt0,$ there exists some $M\gt0$ large enough that
$$\{g\ge M\}\subseteq\{g\ln^+\ln^+g\ge M'\}.$$ And by hypothesis $$\int_{g\ln^+\ln^+g\ge M'}g\ln^+\ln^+g\lt\infty.$$
Hence we can choose $M$ more large enough that $\ln^+\ln^+g\ge1$ when $g\ge M.$ Thus
$$\int_{g\ge M}g\le\int_{g\ge M}g\ln^+\ln^+g\le\int_{g\ln^+\ln^+g\ge M'}g\ln^+\ln^+g\lt\infty.$$
Since $\mu(X)=1,$we conclude that
$$\int_Xg=\int_{g\ge M}g+\int_{g\lt M}g\le \int_{g\ge M}g+M\mu(X)\lt\infty.$$