Let $E\subset\mathbb{R}$ be measurable and for each $n\in\mathbb{N}$, let $f_n:E\rightarrow\mathbb{R}$ be measurable. For each $\varepsilon > 0$, define $E_n(\varepsilon) = \{x\in E : |f_n(x)|\geq \varepsilon\}$. Show tht if $\sum_{n=1}^{\infty}m(E_n(\varepsilon))<\infty$ for every $\varepsilon >0$, then $\lim_{n\rightarrow\infty}f_n(x)=0$ for almost every $x\in E$.

So I've been thinking about this problem for a while now and I just don't see how it even makes sense. What should my set of measure zero be that I'm removing from $E$?

By going far enough down the sequence $\{E_n(\varepsilon)\}$, and then taking the union of the remaining sets in its tail: $\cup_{n=k}^{\infty}E_n$, I can make the measure of this union as small as I like, but ultimately it is fixed, and must be removed from $E$ prior to taking the $\lim_{n\rightarrow\infty}f_n(x)$, and thus it will never necessarily have measure zero.

The only reasonable sets of measure zero I see would be some infinite intersection of sets in the sequence $\{E_n(\varepsilon)\}$. But this seems to make my set of measure zero too restrictive, and I see no way of doing this and still avoiding having points in my remaining set which map under $f_n$, for an infinite number of $n$, to values whose magnitude is greater than $\varepsilon$.

Am I missing something? Thanks.


The idea here is we want to remove some small set and have $f_n$ converge outside of that set.

So we think about what it means to $\mathbf{not}$ converge. This means that for some small $\epsilon > 0$, for every $K$ we have some $n\geq K$ so that $\left|f_n(x)\right| > \epsilon$.

So we have to convert this statement into something we can use in terms of our $E_n(\epsilon)$ sets. First thing is to replace $\epsilon$ with some sort of countable thing, say $\frac{1}{N}$, then we can rewrite this as:

There is some $N$ so that for every $K$ we have some $n\geq K$ so that $\left|f_n(x)\right| > \frac{1}{N}$.

Now this statement is nice because it translates nicely into our $E_n(\epsilon)$ sets. This is the same thing as $f_n$ does not converge at $x$ if $$x \in \bigcup_{N=1}^{\infty} \bigcap_{K=1}^{\infty} \bigcup_{n=K}^{\infty} E_n(\frac{1}{N}) = E.$$

So it suffices to show that $E$ is a small set. To do this, it suffices to show that for each $N$ $\bigcap_{K=1}^{\infty}\bigcup_{n=K}^{\infty} E_n(\frac{1}{N})$ is a null set, because the countable union of null sets is again a null set.

But this is immediate because $$\mu\left(\bigcap_{K=1}^{\infty} \bigcup_{n=K}^{\infty} E_n(\frac{1}{N})\right) \leq \mu\left(\bigcup_{n=K}^{\infty} E_n(\frac{1}{N})\right)$$ for each $K$, so fix $\epsilon > 0$ and since $\sum_1^{\infty} \mu\left(E_n(\frac{1}{N})\right) < \infty$ we can choose some $K$ large so that $\sum_K^{\infty} \mu\left(E_n(\frac{1}{N})\right) < \epsilon$. And so we have $$\mu\left(\bigcup_{n=K}^{\infty} E_n(\frac{1}{N})\right)\leq \sum_K^{\infty} \mu\left(E_n(\frac{1}{N})\right) \leq \epsilon.$$

So we see that $E$ is a null set, and we're done.

  • $\begingroup$ Thanks! this makes perfect sense. $\endgroup$ – Thoth Oct 29 '12 at 3:18

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