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I am studying for a qualifying exam and am having difficulty with this problem:

Let $\left( X, \mathcal{M}, \mu \right)$ be a measure space and assume $f_n \geq 0$ such that $\int f_n = 1$ for all $n$. Show that $$\limsup_n \left( f_n(x) \right)^{\frac{1}{n}} \leq 1 \text{ for a.e. } x. \;\;\;\;\;\;\;\;\;\; (*)$$

Attempts: (a) Notice that if $x$ does not satisfy $(*)$, then by the Root Test $\sum f_n(x) = \infty$, hence if $x$ does not satisfy $(*)$ on a set of positive measure $E$, then $\sum \int_E f_n = \int_E \sum f_n = \infty$. However, this does not seem to contradict the hypothesis.

(b) If $\limsup_n \left( f(x) \right)^{1/n} > 1$, then $\limsup_n f_n(x) = \infty$. Hence, if $x$ does not satisfy $(*)$ on a set of positive measure $E$, then $\infty = \int_E \limsup f_n \geq \limsup \int_E f_n$. Again, this does not seem to contradict the hypothesis.

Thanks in advance for your help.

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  • $\begingroup$ Doug, regarding your idea in item b. Let $C=\{x\in X : \limsup_n \left( f(x) \right)^{1/n} > 1\}$. Since, if $\limsup_n \left( f(x) \right)^{1/n} > 1$ then $\limsup_n f_n(x) = \infty$, we have $$C \subseteq \{x\in X : \limsup_n f_n(x) = \infty\}$$ HOWEVER, we may have $\{x\in X : \limsup_n f_n(x) = \infty\}= X$. So, this path probably will not lead to a solution. $\endgroup$ – Ramiro Feb 4 '16 at 15:01
  • $\begingroup$ I apologize for my (lack of) short term memory, but Ramiro, did you leave a solution yesterday and then take it down? If so, where was the error? It seemed ok when I went through it. $\endgroup$ – Doug Feb 4 '16 at 21:35
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Your idea that if the conclusion fails, then $\sum f_n = \infty$ on a set of positive measure is almost what we need to conclude.

Indeed, consider the modified sequence $g_n := f_n / n^2$. Then $$ \int \sum_n g_n \, d\mu = \sum_n \int g_n \, d\mu = \sum_n 1/n^2 < \infty, $$ so that we get $\sum_n g_n < \infty$ almost everywhere.

But by the root test, this implies $$ 1 \geq \limsup_n \sqrt[n]{g_n(x)} = \limsup_n \frac{(f_n(x))^{1/n}}{\sqrt[n]{n^2}} = \limsup_n (f_n(x))^{1/n} $$ for almost every $x$.

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Alternative Solution:

Take any $\epsilon>0$. We will show that $\mu\{x: \limsup_n (f_n(x))^{\frac{1}{n}}>1+\epsilon\}=0$. Define $A_n=\{x: f_n(x)> (1+\epsilon)^n\}$, and $A=\{x: \limsup_n (f_n(x))^{\frac{1}{n}}>1+\epsilon\}$. Then, $A=\limsup_n A_n$. Now, $$\mu(A_n)=\int_X\mathbb{I}\{A_n\}\le\int_X\frac{f_n(x)}{(1+\epsilon)^n}=(1+\epsilon)^{-n}$$. Therefore, $\sum_{n=1}^{\infty}\mu(A_n)\le\sum_{n=1}^{\infty}(1+\epsilon)^{-n}<\infty$. Hence, by Borel-Cantelli Lemma, $\mu(A)=0$, which proves that $\limsup_n (f_n(x))^{\frac{1}{n}}\le 1$ a.e. $\square$

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  • $\begingroup$ Thanks, this is also a nice solution. I only saw the Borel-Cantelli lemma for the first time this week, but I get the feeling that it is very useful! $\endgroup$ – Doug Feb 4 '16 at 22:07

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