The following is from a Sample Exam question I am studying from, and the question has stumped me.
$$f\in L^{1}[0,1]$$
$$\lim_{n\to\infty}\int_{0}^{1}|f(x)|^{\frac{1}{n}}dx = m(\left\{ {x:f(x)\neq 0}\right\} )$$
What I think:from below: I set $A=\{x:f(x)\neq 0\} ) $ Then $\lim_{n\to\infty}\int_{A}|f(x)|^{\frac{1}{n}}dx \leq \lim_{n\to\infty}\int_{0}^{1}|f(x)|^{\frac{1}{n}}dx$
When I take the limit both sides I get that required lower bound.
from above I get $$\lim_{n\to\infty}\int_{0}^{1}|f(x)|^{\frac{1}{n}}dx \leq 1$$
any insight would be appreciated. My Texing skills are not great, but the above sums up the best of my attempts so far. Thank you
Note that for $n>1$ we have for $f(x)\ne 0$
$$ |f(x)|^{1/n}\le \begin{cases} |f(x)|&,|f(x)|\ge 1\\\\ 1&,|f(x)|<1 \end{cases}$$
Since $|f(x)|^{1/n}\le \max(1,|f(x)|)$, and since $f\in L^1[0,1]$ the Dominated Convergence Theorem guarantees that
$$\lim_{n\to \infty}\int_0^1 |f(x)|^{1/n}\,dx=\int_0^1 \lim_{n\to \infty}|f(x)|^{1/n}\,dx=m\{x: f(x)\ne 0\}$$