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Suppose we have the measure space $[0,1]$ with the lebesgue measure, $\lambda$ on $[0,1]$. Define $g:[0,1] \rightarrow \mathbb{R}$ is measurable and that $g(x) > 0 $ almost everywhere on $[0,1]$. In addition, assume that $g$ is Lebesgue Integrable. Prove for all $n \geq 1$, $g^{1/n}$ is Lebesgue Integrable.

Since $g(x) > 0$ , and is integrable, $g$ is an upper function. Therefore, there is a sequence of step functions, $\{ \phi_m \}$, such that $0 < \phi_m \uparrow g$. Now I want to say that $\phi_{m} ^{\frac{1}{n}}\uparrow g^{1/n}$ for a fixed n. However, I do not think that is true. Also, I really want to say that $\sqrt[n]{g} < g$ almost everywhere on $[0,1]$ and I don't know if that's true either. If that is true, then I could say that $\sqrt[n]{g}$ is Lebesgue Integrable. Do you think I am on the right track? Thank you very much for your help!

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HINT: the inequality $\sqrt[n]{g(x)}<g(x)$ is not true when $g(x)\in[0,1]$. What happen is that $\sqrt[n]{g(x)}< g(x)$ when $g(x)> 1$, and $\sqrt[n]{g(x)}\geqslant g(x)$ when $g(x)\in [0,1]$.

Now define $A:=\{x\in [0,1]:g(x)\leqslant 1\}$ and $B:=\{x\in [0,1]:g(x)> 1\}$ and note that $A \cap B=\emptyset $ and $A \cup B=[0,1]$. Can you finish from here?

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  • $\begingroup$ I'm still confused. Well can we say that the $g$ is integrable on $A \cup B$? $\endgroup$
    – suunySide
    Commented May 1, 2020 at 4:07
  • $\begingroup$ @suunySide $g$ is integrable by asumption, and $A$ and $B$ are measurable by definition (they are Borel sets), then for any chosen measurable function $f:[0,1]\to {\mathbb R}$ $$\int_{[0,1]}f=\int_A f+\int_B f$$ $\endgroup$
    – Masacroso
    Commented May 1, 2020 at 4:13
  • $\begingroup$ I see. Also, I notice that one of the sets is a complement of the other set. So could we cay that $\int_{[0,1]} \sqrt[n]{g} = \int_{A} \sqrt[n](g) + \int_{B} \sqrt[n]{g}$ ? $\endgroup$
    – suunySide
    Commented May 3, 2020 at 16:23

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