# Prove that $\limsup_{n \rightarrow \infty} (f_n(x))^{1/n} \leq1$

Suppose $f_n$ are nonnegative measurable functions on a measure space satisfying $$\int f_n\ \mathsf d\mu =1.$$ Prove that $$\limsup_{n \rightarrow \infty} (f_n(x))^{1/n} \leq1$$ almost everywhere.

I'm considering to use either by contradiction or Borel- Cantelli lemma. For the first case, I assumed there exists an $n$ such that the conclusion is not true and then we have a contradiction with the integral being 1. But I think this should use Borel-Cantelli lemma which I'm not sure how to apply.

If $\int f_n = 1$ for all $n,$ then

$$\sum (1/n^2)\int f_n = \sum \int f_n/n^2 = \int (\sum f_n/n^2) < \infty,$$

where we used the monotone convergence theorem to get the second equality. Thus $\sum f_n(x)/n^2 < \infty$ for a.e. $x.$ Fix such an $x.$ Because the series converges, we have

$$f_n(x)/n^2 \to 0\implies f_n(x)/n^2 < 1 \ \text {for large}\ n \implies f_n(x)^{1/n} < (n^2)^{1/n} \ \text {for large}\ n .$$

Now take the $\limsup,$ recalling $(n^2)^{1/n} \to 1.$

• It's hard for me to come up with $f_n /n^2$ though. – hil316 Aug 9 '15 at 6:20
• I'm not sure what you are saying there. – zhw. Aug 10 '15 at 5:01

Hints:

1. Fix $\epsilon>0$. Show using Markov's inequality that $$\mu(\{x; f_n(x)^{1/n} \geq 1+\epsilon\}) \leq \frac{1}{(1+\epsilon)^n}.$$
2. Conclude from the Borel-Cantelli theorem that $$f_n(x)^{1/n} \leq 1+\epsilon$$ for all $n \geq N(x)$ sufficiently large for almost all $x$.
3. Conclude that $$\limsup_{n \to \infty} f_n(x)^{1/n} \leq 1+\epsilon$$ almost everywhere. Since $\epsilon>0$ is arbitrary, the claim follows.
• I probably miss something, but it's it enough to notice that $f_n<\alpha <1$ due to the integral value for some constant $\alpha$, and hence the limit is 1. – Alex Aug 8 '15 at 21:35
• @Alex $\int f_n \, d\mu = 1$ does not imply $f_n<\alpha<1$; just consider e.g. $f_n(x) := n 1_{[0,1/n]}(x)$. – saz Aug 9 '15 at 5:46
• Thanks, so the support is $[0,\frac{1}{n}]$. I guess I missed it. – Alex Aug 9 '15 at 8:42