Riemann rearrangement theorem for $L^1$ functions

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.

• 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 $f$. Apr 28 at 7:10
• $A_n$ are chosen freely, so there should be no constraints. Apr 28 at 7:19
• I don't think this works when $\delta_n = 1$. Apr 28 at 8:10
• 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 $\sum |c_n| \delta_n = \infty$.
– mlk
Apr 28 at 8:31
• 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. 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.

• $f-f_k$ 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.
– Eric
Apr 29 at 12:52
• @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 $\sup_{[\epsilon,1]} f$.
– mlk
Apr 29 at 19:43
• Thanks for clarifying - neat solution!
– Eric
Apr 29 at 21:29