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 $\int g\ {\rm d} \mu<\infty$?
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1what does $\ln^+$ mean? – user190080 Jul 18 '16 at 15:17
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Maybe $$\ln^+ x=\ln \mid x\mid?$$ – awllower Jul 18 '16 at 15:24
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1It means: $\ln^+ x = \max (\ln x,0)$ – LeeWoo Jul 18 '16 at 15:34
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2and how do you handle expressions like $\ln(0)$ - which can clearly happen? – user190080 Jul 18 '16 at 16:08
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1I guess it will be like this $\ln^+ 0 = \max (\ln 0,0):=0$, so you don't need to care about what $\ln 0$ is it. – yoyo Jul 18 '16 at 16:35
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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
$$\int_Xg=\int_{g\ge M}g+\int_{g\lt M}g\le \int_{g\ge M}g+M\mu(X)\lt\infty.$$
yoyo
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3Can I choose $M=e^e$? Then we have: $$ \int_{g\ge e^e} g \le \int_{g \ge e^e} g \ln^+ \ln^+ g <\infty $$ and hence $$ \int_{X} g = \int_{g\ge e^e} g + \int_{g\le e^e} g \le \int_{g\ge e^e} g +e^e \mu(X)<\infty. $$ – LeeWoo Jul 18 '16 at 18:19
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