# Can you help on this question on Lebesgue Integrals

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!

HINT: the inequality $$\sqrt[n]{g(x)} is not true when $$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]$$. Can you finish from here?
• I'm still confused. Well can we say that the $g$ is integrable on $A \cup B$? Commented May 1, 2020 at 4:07
• @suunySide $g$ is integrable by asumption, and $A$ and $B$ are measurable by definition (they are Borel sets), then for any chosen measurable function $f:[0,1]\to {\mathbb R}$ $$\int_{[0,1]}f=\int_A f+\int_B f$$ Commented May 1, 2020 at 4:13
• I see. Also, I notice that one of the sets is a complement of the other set. So could we cay that $\int_{[0,1]} \sqrt[n]{g} = \int_{A} \sqrt[n](g) + \int_{B} \sqrt[n]{g}$ ? Commented May 3, 2020 at 16:23