Let $f_n$ be a sequence of non-negative functions in $L^1(X,\mu)$ such that $\|f_n\|_1=1$ for all $n$. Show that $\limsup_n f_n^{1 \over n}\leq1$ ae. $[\mu]$.
My failed try is to assume on the contrary, $\mu(\cap_n\cup_{k\geq n} \{f_k^{1/k}>1\})=a>0$ $\mu(\cup_{k\geq n} \{f_k^{1/k}>1\})\geq a$. I want to integrate $f_n$ on this set. But no contradiction can be drawn.
On
By countable additivity it's enough to show that if $\alpha>1$ then $\limsup f_n^{1/n}\le\alpha$ almost everywhere. Say $E$ is the set where this limsup is greater than $\alpha$. Then $$E\subset\bigcap_n\bigcup_{k\ge n}E_{\alpha,k},$$where $$E_{\alpha,k}=\{f_k>\alpha^k\}.$$Now $\int f_k=1$ says that $\mu(E_{\alpha,k})$is less than something, and it follows that $$\lim_n\sum_{k\ge n}\mu(E_{\alpha,k})=0.$$
By Markov's inequality
$$\mu \left( \bigcup_{k \geq n} \{f_k^{1/k}>(1+\epsilon)\} \right) \leq \sum_{k \geq n} \mu(f_k>(1+\epsilon)^k) \leq \sum_{k \geq n} \frac{1}{(1+\epsilon)^k} \underbrace{\int f_k \, d\mu}_{1}$$
which implies
$$\mu \left( \limsup_{n \to \infty} f_n^{1/n}>1+\epsilon \right) = \lim_{n \to \infty} \mu \left( \bigcup_{k \geq n} \{f_k^{1/k}>(1+\epsilon)\} \right) = 0$$
for any $\epsilon>0$. As $\epsilon>0$ is arbitrary, this proves
$$\mu \left( \limsup_{n \to \infty} f_n^{1/n}>1 \right)=0.$$