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Problem

Let $\{f_k\}_{k \in \mathbb N}$ be a sequence of measurable functions defined on $E \subset \mathbb R^n$ with $E$ measurable and $|E|<\infty$, such that $f_k \to 0$ a.e.. Show that there exists a subsequence $\{f_{k_j}\}_{j \in \mathbb N}$ such that $\sum_{j \in \mathbb N} |f_{k_j}|<\infty$ a.e.

I thought of using Egorov's theorem, so for each $\epsilon_n=\dfrac{1}{n}$, there exists a closed subset $F_n \subset E$ with $|E \setminus F_n|<\epsilon_n$ and such that $f_n \rightrightarrows 0$ on $F_n$. For each $j$, I can pick $n_j$ with $|f_{n_j}|<\dfrac{1}{2^j}$ in $F_j$ and I can also pick $n_1<n_2<...<n_j<...$

It is easy to see that complement of the set $F=\bigcup_{j \in \mathbb N} F_j$ has measure zero. The problem is that I cannot affirm $|f_{n_j}|<\dfrac{1}{2^j}$ in all $F$ but just in $F_{n_j}$. I can assure this on the intersection $\bigcap_{j \in \mathbb N} F_j$ but the complement of this set is not of measure zero, so the series is not convergent almost everywhere.

Any hints to solve this problem would be greatly appreciated.

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2 Answers 2

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Construct $F_n$ as you did, but then let $F_n' = F_1 \cup \dots \cup F_n$. Then we again have $|E \setminus F_n'| < \epsilon_n$ and $f_n \rightrightarrows 0$ on $F_n'$. Moreover, $F_1' \subset F_2' \subset \dots$ which will be useful later. Choose the $n_j$s such that $|f_{n_j}| < 2^{-j}$ on $F_{j}'$ instead. Now let's set $F = \bigcup_n F_n'$ instead.

Hint 2: If you do that, then for any fixed $x \in F$, you cannot guarantee $|f_{n_j}(x)| < \frac{1}{2^j}$ for every $j$, but you can guarantee it for all but finitely many $j$...

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  • $\begingroup$ Thanks for your answer, I have some doubts: suppose I could construct these nested sets, for each $j$, I pick $n_j$ with $|f_{n_j}| \leq \dfrac{1}{2^j}$ and I can pick $n_1<...<n_k<...$. $\endgroup$
    – user16924
    Commented Sep 12, 2015 at 0:36
  • $\begingroup$ Since you've mentioned "all but finitely many $j$", I suppose that your suggestion is to consider the set $F=\bigcup_{k \in \mathbb N} \bigcap_{n\geq k} F_n$. Then, for each $x \in F$, there is $k$ such that $|f_{n_j}(x)|<\dfrac{1}{2^j}$ for all $j \geq k$., so we would have $\sum_{j=1}^{\infty}|f_{n_j}(x)|=\sum_{j=1}^{k-1}|f_{n_j}(x)|+\sum_{j=k}^{\infty}|f_{n_j}(x)|<\sum_{j=1}^{k-1}|f_{n_j}(x)|+\sum_{j=k}^{\infty}\dfrac{1}{2^j}< \infty$. I don't see how can I choose the closed sets $F_n$ such that they form an increasing sequence, could you help me with that? $\endgroup$
    – user16924
    Commented Sep 12, 2015 at 0:36
  • $\begingroup$ That's not what I meant - if you define $F$ that way, you can no longer guarantee that $F^c$ has measure 0. Rather, see my edit. $\endgroup$ Commented Sep 12, 2015 at 1:41
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Convergence a.e. on a finite measure space implies convergence in measure. So we may choose $k_1$ such that $k\geqslant k_1$ implies $\mu\left(\left\{x:|f_k(x)|> 2^{-k}\mu(E)^{-1}\right\}\right)<2^{-k}\mu(E)^{-1}$, and inductively $k_j\geqslant k_{j-1}$ such that $k\geqslant k_j$ implies $\mu\left(\left\{x:|f_k(x)|>2^{-k_j}\mu(E)^{-1}\right\}\right)<2^{-k_j}\mu(E)^{-1}$ and $k_j\stackrel{j\to\infty}\longrightarrow\infty$. We may assume WLOG that $2^{-k_1}\mu(E)^{-1}<1$. Hence \begin{align} \sum_{j=1}^\infty |f_{k_j}| &\leqslant \sum_{j=1}^\infty 2^{-k_j}\mu(E)^{-1}\mu(E)+2^{-k_j}\mu(E)^{-1}\\ &\leqslant \mu(E)^{-1}\sum_{j=1}^\infty 2^{-k_j}\\ &\leqslant \mu(E)^{-1}\\ &<\infty. \end{align}

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  • $\begingroup$ P.S. I'm really tired so if my proof is flawed feel free to point it out and I will correct it later. $\endgroup$
    – Math1000
    Commented Sep 11, 2015 at 1:19
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    $\begingroup$ Unless there are some a.e.'s missing, you seem to have "proved" that $\sum_j |f_{k_j}| < \infty$ everywhere, which cannot be generally true (suppose there is some fixed null set where every $f_k$ equals 1). $\endgroup$ Commented Sep 11, 2015 at 5:02

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