Let $c_n$ be a sequence of real numbers with $\sum c_n$ converging conditionally but not absolutely. Suppose $\delta_n > 0$ is another sequence with $\delta_n \to 0$, and $\sum c_n \delta_n$ converging also conditionally but not absolutely.

Does there exist, for every $L^1$ function $f: [0, 1] \to \mathbb R$, a bijection $\gamma: \mathbb N \to \mathbb N$, and a sequence of measurable sets $A_n$ with $\mu(A_n) = \delta_n$ such that

$\sum c_{\gamma(n)}1_{A_{\gamma(n)}} \to f$,

in $L^1$ and pointwise a.e?

Note: Here $\mu$ denotes the Lebesgue measure.

  • $\begingroup$ Are you missing some constraints on $A_n$? As is you can take them to all be the same set, or to be disjoint from the support of $f$. $\endgroup$ Apr 28 at 7:10
  • $\begingroup$ $A_n$ are chosen freely, so there should be no constraints. $\endgroup$
    – Nate River
    Apr 28 at 7:19
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    $\begingroup$ I don't think this works when $\delta_n = 1$. $\endgroup$
    – Leo Moos
    Apr 28 at 8:10
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    $\begingroup$ If you pick $|c_n| = \delta_n = 1/n$, then you can estimate $\int | \sum c_{\gamma(n)} 1_{A_{\gamma(n)}} | dx \leq \sum |c_n| \delta_n = \sum \frac{1}{n^2} < \infty$, so to approximate arbitrary functions, you might need to also require that $\sum |c_n| \delta_n = \infty$. $\endgroup$
    – mlk
    Apr 28 at 8:31
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    $\begingroup$ As stated I don't think it would work. For example, let $c_n$ be the alternating harmonic sequence. Let $\delta_n = \frac{1}{\ln(n+3)}$ if $n$ is odd, and $\frac{1}{n}$ if $n$ is even. Then you cannot use this to approximate any $f$ whose negative part has mass greater than 10. $\endgroup$ Apr 28 at 19:16

The problem is trickier than I initially thought, but with the corrected condition it can be done. I need to assume that $\sum c_n \delta_n$ is conditionally but not absolutely convergent, $0 < \delta_n < 1$ with $\delta_n \to 0$ and $c_n \to 0$ (which is currently only implied by the conditional convergence of $\sum c_n$).

Wlog. we can assume $f: [0,1] \to \mathbb{R}$ to be non-negative (otherwise split the series in two and use them to approximate positive and negative part separately) and non-increasing (by rearrangement, just to simplify notation).

Now proceed as following: Denote the current partial sum by $\tilde{f}_k$. Take the next unused $n\in \mathbb{N}$ such that $c_n > 0$ and consider the sets $$A_\lambda := \{x \in [\lambda,1]: f(x)- \tilde{f}_k(x) \geq c_n \}$$ for $\lambda \in [0,1]$. If there is a $\lambda$ such that $|A_\lambda| = \delta_n$, then choose $\gamma(k) := n$, $A_n := A_\lambda$, add $c_n 1_{A_n}$ to the partial sum and iterate. (The "positive process")

If not, take the next unused $m \in \mathbb{N}$ such that $c_m < 0$, choose $\gamma(k) := m$ and find a set $A_m \subset [0,2\max(\delta_n,\delta_m)]$ with $|A_m| =\delta_m$, for which the approximation is best, i.e. $x \notin A_m$ implies $f(x)-\tilde{f}_k(x) \geq f(y)- \tilde{f}_k(y)$ for all $y \in A_m$, add $c_m 1_{A_m}$ to the partial sum and iterate. (The "negative process")

The positive process cannot continue indefinitely, since it increases the integral by $c_n \delta_n$ and $\tilde{f}_k \leq f$. The negative process cannot continue indefinitely, since at some point it will achieve $f(x) - \tilde{f}_k(x) \geq c_n$ on $[0,\delta_n]$, at which point the positive process will take over again. Thus we use all indices and have constructed a bijection.

Now fix $\epsilon > 0$. At some point, all future $\delta$ are smaller than $\epsilon/2$, so after that on the interval $[\epsilon,1]$ the approximation $\tilde{f}_k$ is monotone increasing in $k$. Furthermore, whenever the negative process takes over, we know that $f(x) - \tilde{f}_k < c_n$ except for a set of measure $< \delta_n$. Thus as $f$ is decreasing, $$\int_\epsilon^1 f - \tilde{f}_k dx < c_n (1-\epsilon) + \delta_n \sup_{[\epsilon,1]} (f-\tilde{f}_k).$$ Here, the first term converges to $0$ as $c_n \to 0$, while for the second term $\sup_{[\epsilon,1]} f \leq f(\epsilon)$ as $f$ is non-increasing and $\sup_{[\epsilon,1]} (- \tilde{f}_k)$ is bounded as there were only a finite number of negative $c_m$ until we reached monotonicity on $[\epsilon,1]$. Then since $\delta_n \to 0$, that term converges as well and we have that $\tilde{f}_k \to f$ in norm on $[\epsilon,1]$, which together with monotonicity implies pointwise convergence a.e. on $[\epsilon,1]$ and as $\epsilon$ was arbitrary, we then have pointwise convergence a.e. on $[0,1]$.

Finally consider the negative process. We know that $|\{x\in [0,2 \max(\delta_n,\delta_m)]: \tilde{f}_k(x) < -c_n \}| <\delta_n \leq \max(\delta_n,\delta_m)$, otherwise the positive process would already have taken over. But this means that for the set $A_m$ we choose, we have $\tilde{f}_k(x) \geq -c_n$ for all $x\in A_m$ and hence at all points modified, we will have $\tilde{f}_{k+1} (x) \geq -c_n +c_m$. But since this is true for all negative steps, we have $\tilde{f}_k \geq - 2\max_{n\in \mathbb{N}} |c_n|$ for all $k\in\mathbb{N}$. But than we have an integrable lower bound which allows us to conclude that the pointwise limit $f$ is also the $L^1$ limit of $\tilde{f}_k$ by dominated convergence.

  • $\begingroup$ $f-f_k$ doesn’t stay monotonic. Your negative processes are all subsets of a neighborhood of 0 while your positive additions are subsets of a neighborhood of 1, so I don’t understand how repeated negatives enable a positive. Are you resorting on each new value? If so, that seems to break your assumptions in the latter part of the proof. $\endgroup$
    – Eric
    Apr 29 at 12:52
  • $\begingroup$ @Eric Since the positive process takes precedence, $\lambda$ will get close to $0$ after a while, so there will be positive additions even in the interval touched by the negative process. In fact, most of the additions will happen there. And I never claimed or used that $f-f_k$ is monotonic, only that $f$ is and only as a quick way to estimate $\sup_{[\epsilon,1]} f$. $\endgroup$
    – mlk
    Apr 29 at 19:43
  • $\begingroup$ Thanks for clarifying - neat solution! $\endgroup$
    – Eric
    Apr 29 at 21:29

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