I am trying to solve this problem:
Let $f \in L^1([0,1])$ be a non negative, finite function. Show that $$\lim_{n \to \infty} \int_0^1 \sqrt[n]{f(x)}dx=m(\{x \in [0,1]/f(x)>0\}$$
This is what I could do:
Let $f_n(x)=\sqrt[n]{f(x)}$. If $x \in [0,1]$ such that $f(x)=0$, then $f_n(x)=0$ for all $n$, if $f(x)>0$, then $f_n(x) \to 1$. So $f_n(x) \to \mathcal X_S$, where $S=\{x \in [0,1]: f(x)>0\}$.
Since $m(\{x \in [0,1]: f(x)>0\})=\int_0^1\mathcal X_Sdx$, then all I have to show is $$\lim_{n \to \infty} \int_0^1 f_ndx=\int_0^1 \lim_{n \to \infty}f_ndx$$
I don't know which property allows me to affirm that the limit of the integrals is the integral of the limit, I am not under the hypothesis of the monotone convergence theorem ($f_n(x)$ is increasing for $x:f(x) \geq 1$ and decreasing if not), I cannot apply the dominated convergence theorem since from the information of the problem I can't assure there exists $g$ integrable with $|f_n|\leq |g|$).
Any suggestions would be appreciated.
For $x\in[0,1]$, define $$ g(x):=\max\{f(x), 1\}. $$ Then $\lvert f_n(x)\rvert\leq g(x)$ for all $x\in[0,1]$: we have equality if $f(x)\geq1$, and if $f(x)\leq 1$ then $\sqrt[n]{f(x)}\leq 1$ for all $n$.
Can you prove that $g(x)$ is integrable? If so, then the Dominated Convergence Theorem will do it for you.