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 $\sqrt[n]{g(x)}<g(x)$ 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?