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

Question

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.

Answer 1

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 $(n^2)^{1/n} \to 1.$

Answer 2

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.