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Let $g:X \rightarrow \mathbb{R}$ be a measurable function on the measure space $(X,A,\mu)$ where $\mu(X)< \infty$. If $g$ is Lebesgue integrable, how can I show $[log(1+|g(x)|)]^8$ is Lebesgue integrable?

I most likely need to make use of the comparison test but I’m struggling to use it for this problem.

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Let $X=(0,1)$ and $g(x) = e^{1 \over \sqrt[8]{x}} -1$. Then $g$ is measurable , $\mu X $ is finite, but $x \mapsto (\log(1+|g(x)|))^8 = { 1\over x}$ is clearly not integrable.

If we let $\phi(x) = (\log (1+x))^8$ on $[0,\infty]$ it is not too difficult to show that $|\phi(x)| \le L=6588344 \ e^{-7}$, in particular $\phi$ is Lipschitz with rank $L$.

Hence $|\phi(g(x))| \le |\phi(g(0)) | + L |g(x)| + L|g(0)|$ and so $\int | \phi \circ g | \le \mu X (|\phi(g(0)) | + L |g(0)|) + L\int |g|$, hence $\phi \circ g$ is integrable.

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  • $\begingroup$ I apologise. I realise I accidentally said lebesgue measurable instead of lebesgue integrable as the condition on $g$. $\endgroup$
    – Partey5
    Commented Oct 27, 2020 at 21:21
  • $\begingroup$ I updated my answer. $\endgroup$
    – copper.hat
    Commented Oct 28, 2020 at 22:04

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