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**Question from an exam sample I’m studying for:** Suppose $\left(X,\mathcal{F},\mu\right)$

is a measure space and $f_{n}:\left(X,\mathcal{F}\right)\to\mathbb{R}$

a sequence of non-negative integrable functions such that $\int_{X}f_{n}d\mu=1$

. Is is it necessarily true that $\frac{1}{n}f_{n}$

converges almost everywhere to $0$

? what about $\frac{1}{n^{2}}f_{n}$

?

**My line of thought:** From Fatou’s llema we got that: $$\int\limits _{X}\liminf_{n\to\infty}\frac{1}{n}f_{n}d\mu\geq\liminf_{n\to\infty}\int\limits _{X}\frac{1}{n}f_{n}d\mu=\liminf_{n\to\infty}\frac{1}{n}\int\limits _{X}f_{n}d\mu=\liminf_{n\to\infty}\frac{1}{n}=0$$

Thus $g:=\liminf\frac{1}{n}f_{n}$

is a non-negative function whose integral equals zero and thus $g=0$

almost-everywhere. So if I knew that $\frac{1}{n}f_{n}$

converged almost-everywhere then in particular it would be equal to $g$

almost-everywhere and thus to $0$

. I’ve been racking my head trying to think whether it’s possible to construct a sequence with the given properties such that $\frac{1}{n}f_{n}$

would fail to converge on a set of positive measure. I want to say that this is impossible and that both claims are true (which would make the phrasing of the question quite mean since these questions usually imply one case is true and one is false) but I’m not certain.

I’d really appreciate help clearing this up!

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**Edit:** Since you can only accept one answer I wanted to thank Did, Vincent and M. Luethi for their replies. Combined we got two counter-examples for $\frac{1}{n}$ and a proof for $\frac{1}{n^{2}}$.

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The typewriter sequence disproves the almost sure convergence of $\frac1nf_n$ (see Example 4 here). To study $\frac1{n^2}f_n$, generalize slightly the setting and consider the function $g=\sum\limits_na_nf_n$ where $\sum\limits_na_n$ is a series with positive entries and finite sum. Henri Lebesgue taught us that $$\int g=\sum_na_n\int f_n=\sum_na_n$$ is finite hence $g$ is finite almost everywhere, in particular $a_nf_n\to0$ almost everywhere. This applies to $a_n=\frac1{n^2}$.

I make it an answer but I leave out a few details so that you still have to look at it a bit. I can add the details later if need be.

Look at the unit interval $[0,1]$ and define a collection of natural coverings: $$\begin{align}P_{k}=\left\{\left[0,\frac{1}{2^{k}}\right],\ldots,\left[\frac{2^{k}-1}{2^{k}},1\right]\right\}\end{align}$$ Define the sequence $f_{n}$ by choosing $k$ such that $2^{k}\leq n < 2^{k+1}$ and let $f_{n}:=2^{k}1_{\left[\frac{n-2^{k}}{2^{k}},\frac{n+1-2^{k}}{2^{k}}\right]}$.

If $n\in\mathbb{N}$, then $\frac{1}{n}f_{n}(x)=\frac{2^{k}}{n}$ where $2^{k}\leq n\leq 2^{k+1}$, so that $\frac{1}{2}\leq \frac{2^{k}}{n}\leq 1$. In order to see that every $x\in[0,1]$ lands in infinitely many of the intervals, it suffices to check that for all $k\in\mathbb{N}$ and for all $0\leq l<2^{k}$, there exists a natural number $n\in\mathbb{N}$ such that $n=2^{k}+l$, because then all $P_{k}$ appear in our sequence. But the existence of $n$ is obvious.

Define the sets $A_{n}$ as following:

$$

A_{n} = \{ x \in [0,1] \, : \, \mbox{x – y is an integer with} \, \sum_{k=1}^{n-1} \frac{1}{k} \leq y \leq \sum_{k=1}^{n} \frac{1}{k}\}.

$$

Then $\mu(A_{n}) = \frac{1}{n}$. Now define $f_{n}(x) = n \cdot \chi_{A_{n}}$ with $\chi_{A_{n}}$ the indicator function on $A_{n}$. Then $\int_{X} f_{n} d\mu = 1$ follows. Can you prove that $\frac{1}{n} f_{n}(x)$ converges nowhere on $[0,1]$?

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