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I am asking myself this question (which seems to be a natural generalization of Remark 4.4 of these lecture notes).

Question. Let $I_s, s \in \mathcal{S}$ be a collection of intervals included in $[0,1]$ such that there exists $K>0$ such that for every interval $J$ there holds $$\frac{1}{|J|} \sum_{I_s\subset J} |I_s|\leqslant K\,,$$ then is it true that the function $f:=\sum_{s\in\mathcal{S}} \mathbf1_{I_s}$ is in $\bigcap_{p \geqslant 1}L^p$? Is it also true that $|\{x \in [0,1]\,, f(x)\geqslant n\}| \lesssim e^{-cn}$ for some $c>0$?

I have the strong intuition that the assumption on the intervals implies that the function $f$ is not big on large measure sets.

A first guess would be that $f$ lies in $\mathit{BMO}$ (i.e., has bounded mean oscillation) but as pointed out by fedja it is not the case. If we take the family of intervals $[1/2,1/2+2^{-j}]$ we see that it defines a function which is roughly $x \mapsto \log (x-1/2) \mathbf{1}_{x>1/2}$ which is not in $\mathit{BMO}$ (see the averages on $[1/2-\delta, 1/2+\delta]$).

However, if we define $\alpha_n$ as the measure of the $x\in [0,1]$ such that $f(x)=n$ (i.e. $x$ lies in $n$ intervals $I_s$), I thing that we can prove an estimate of the form $\alpha_n \lesssim e^{-cn}$, and for that, it is sufficient to prove that for large $n$ there is a positive real $b<1$ such that $\alpha_{n+1} \leqslant b \alpha_n$.

In order to prove the last statement I tried to select the dyadics intervals $J$ on which $\frac{1}{|J|}\int_J f$ is large or not, and run pretty much the same argument as in the proof of the John-Nirenberg inequality [see Grafakos "Modern Fourier Analysis" page 124]. However the problem is that even if we have: $$\frac{1}{|J|}\int_J f = \frac{1}{|J|}\sum_{\cup I_s \cap J} |I_s|\,,$$ the hypothesis does not say something about this quantity ... (if $I_s$ is not included in $J$ then this interval does not appear in the hypothesis ...).

Can someone give me a hand for proving this claim? I really believe that the proof should be an adaptation of the John-Nirenberg inequality.

Edit. I edited the question after comments.

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    $\begingroup$ Something is fishy: consider the collection of open intervals centered at $1/2$. Then the condition is almost void. Are you sure those are not dyadic intervals or that the condition is not imposed with $J$ being a union of 2 adjacent dyadic intervals? $\endgroup$
    – fedja
    Commented Mar 19, 2020 at 16:21
  • $\begingroup$ Well, I realize that it is not clear. In fact in the lecture notes I just attached a 'tiling' is defined with dyadic intervals so all the intervals involved are in fact dyadic. However we can cahnge the question to make it more interesting by letting $J$ be any interval! $\endgroup$
    – J.Mayol
    Commented Mar 19, 2020 at 18:07
  • $\begingroup$ But then there is something I do not understand with the Remark 4.4 in the attached lecture notes. If any interval involved are dyadic then by Lebesgue differentiation theorem we get that $f$ is bounded and thus in any $L^p$ ... which is strange. $\endgroup$
    – J.Mayol
    Commented Mar 19, 2020 at 18:11
  • $\begingroup$ @fedja I edited the question so that the hypothesis concerns any interval $J$, thus ruling out your example (in your case we can take $J=[1/2-\eta,1/2+\eta]$... $\endgroup$
    – J.Mayol
    Commented Mar 19, 2020 at 18:22
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    $\begingroup$ The BMO understood literally is still out of question: Just consider the family $[0.5,0.5+2^{-j}]$ and look at a very short interval centered at $0.5$. So, you need to be a bit more inventive. Hint: replace each interval by a trapezoid function of comparable width; then the sum will be in BMO, indeed. $\endgroup$
    – fedja
    Commented Mar 19, 2020 at 18:32

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