## Open Question on Sufficiency To Be a Density Basis

This post is adapted from a question I posted on Math.StackExchange approximately two weeks ago, which has not received a satisfactory answer. If you have a solution or a counterexample, please add it to Math.SE instead of here.

Let ${\mathcal{B}}$ be a differentiation basis. Given a measurable set ${E}$, we say that ${\mathcal{V}\subset\mathcal{B}}$ is a Vitali covering of ${A}$ if for every ${x\in E}$ there exists a sequence ${\left\{R_{k}(x)\right\}\subset\mathcal{V}}$ containing and with diameter tending to zero.

Question 1 Suppose that for every measurable set ${A}$ of finite measure, every Vitali covering ${\mathcal{V}}$ of ${A}$, and every ${\varepsilon>0}$, there exists a countable subcollection ${\left\{R_{k}\right\}\subset\mathcal{V}}$ with the following properties:

1. (Underflow condition) ${\left|A\setminus\bigcup R_{k}\right|=0}$
2. (Overflow condition) ${\left|\bigcup R_{k}\setminus A\right|\leq\varepsilon}$
3. (Overlap condition) ${\left\|\sum\chi_{R_{k}}\right\|_{L^{1}}\leq C\left|A\right|}$

where ${C}$ is a positive constant independent of ${A}$, ${\mathcal{V}}$, and ${\varepsilon}$. Does it follow that ${\mathcal{B}}$ satisfies the density property

$\displaystyle \lim_{k\rightarrow\infty}\dfrac{\left|A\cap R_{k}\right|}{\left|R_{k}\right|}=\chi_{A}(x) \ \text{ a.e.} (*)$

where ${x\in R_{k}\in\mathcal{B}}$ for all ${k}$ and the diameter of ${R_{k}}$ tends to zero.

My motivation for posing this question is that when ${q>1}$ and ${1/p+1/q=1}$, a variant of this theorem holds when replace the overlap condition above with ${\left\|\sum\chi_{R_{k}}\right\|_{L^{q}}\leq C\left|A\right|^{1/q}}$ and (*) by

$\displaystyle \lim_{k\rightarrow\infty}\dfrac{1}{\left|R_{k}\right|}\int_{R_{k}}f=f(x) \ \text{ a.e.},$

for every sequence ${\left\{R_{k}\right\}}$ as above and ${f\in L_{loc}^{p}}$. (There is no issue in considering only simple functions above, as opposed to the entire space ${L_{loc}^{\infty}}$, since the differentiaton of the integrals of each space are equivalent.) In fact, the above three conditions are a special case of what is called the ${V_{q}}$ (${1\leq q\leq\infty}$) covering property:

Definition 1 A differentiation basis ${\mathcal{B}}$ has the ${V_{q}}$ covering property if for every measurable set ${A}$ with ${\left|A\right|<\infty}$, every Vitali covering ${\mathcal{V}}$ of ${A}$, and every ${\varepsilon>0}$, there exists a countable subcollection ${\left\{R_{k}\right\}\subset\mathcal{V}}$ satisfying

1. ${\left|A\setminus\bigcup R_{k}\right|=0}$
2. ${\left|\bigcup R_{k}\setminus A\right|\leq\varepsilon}$
3. ${\left\|\sum\chi_{R_{k}}\right\|_{L^{q}}\leq C\left|A\right|^{1/q}}$

where ${C}$ is a constant independent of ${A}$, ${\mathcal{V}}$, and ${\varepsilon}$.

Theorem 2 If ${\mathcal{B}}$ satisfies the ${V_{q}}$ property, ${1, then ${\mathcal{B}}$ differentiates the integrals of functions in ${L_{loc}^{p}({{\mathbb R}}^{n})}$.

Briefly we introduce the notation

$\displaystyle \overline{D}(\int f,x):=\sup_{\left\{R_{k}\right\}}\limsup_{k\rightarrow\infty}\dfrac{1}{\left|R_{k}\right|}\int_{R_{k}}f \ \underline{D}(\int f,x):=\inf_{\left\{R_{k}\right\}}\liminf_{k\rightarrow\infty}\dfrac{1}{\left|R_{k}\right|}\int_{R_{k}}f$

where the supremum and infimum are taken over all sequences in ${\left\{R_{k}\right\}}$ containing ${x}$ and with diameter tending to zero. When both ${\overline{D}(\int f,x)}$ and ${\underline{D}(\int f,x)}$ are equal, we simply write ${D(\int f,x)}$.

Proof: It suffices to show that the set

$\displaystyle E_{\alpha}:=\left\{x\in{{\mathbb R}}^{n}:\left|\overline{D}(\int f, x)-f(x)\right|>\alpha\right\}$

has measue zero for all ${\alpha>0}$. Without loss of generality we may assume that ${f}$ has compact support. Let ${g}$ be a continuous function of compact support, such that ${\left\|f-g\right\|_{L^{p}}<\varepsilon}$. Observe that

$\displaystyle \limsup_{k\rightarrow\infty}\dfrac{1}{\left|R_{k}\right|}\int_{R_{k}}\left|f(y)-f(x)\right|dy\leq\limsup_{k\rightarrow\infty}\dfrac{1}{\left|R_{k}\right|}\int_{R_{k}}\left|f-g\right|+\left|f(x)-g(x)\right|$

So if ${x\in E_{\alpha}}$, then ${x\in F_{\alpha/2}\cup G_{\alpha/2}}$, where

$\displaystyle F_{\alpha/2}:=\left\{x\in{{\mathbb R}}^{n}:\overline{D}(\int\left|f-g\right|,x)>\alpha/2\right\},\indent G_{\alpha/2}:=\left\{x\in{{\mathbb R}}^{n}:\left|f(x)-g(x)\right|>\alpha/2\right\}$

Intersecting ${F_{\alpha/2}}$ with a compact set if necessary, we may assume that ${\left|F_{\alpha/2}\right|<\infty}$. By definition, for each ${x\in F_{\alpha/2}}$ there exists a sequence ${\left\{R_{k}^{x}\right\}\subset\mathcal{B}}$ containing ${x}$ and with diameter tending to zero. The collection of these sets forms a Vitali cover of ${F_{\alpha/2}}$, whence there exists a subcollection ${\left\{R_{k}\right\}}$ satisfying the ${V_{q}}$ conditions. By Hölder’s inequality,

$\displaystyle \begin{array}{lcl}\displaystyle\left|F_{\alpha/2}\right|\leq\sum_{k}\left|R_{k}\right|\leq\dfrac{2}{\alpha}\int_{\bigcup R_{k}}\left|f-g\right|\sum\chi_{R_{k}}&\leq&\displaystyle\dfrac{2}{\alpha}\left\|f-g\right\|_{L^{p}}\left\|\sum\chi_{R_{k}}\right\|_{L^{q}}\\[2 em] &\leq&\displaystyle\dfrac{2C}{\alpha}\varepsilon\left|F_{\alpha/2}\right|^{1/q}\end{array}\$

Dividing both sides of the inequality by ${\left|F_{\alpha/2}\right|^{1/q}}$ gives the desired estimate. By Chebyshev’s inequality,

$\displaystyle \left|G_{\alpha/2}\right|\leq\left(\dfrac{2}{\alpha}\left\|f-g\right\|_{L^{p}}\right)^{p}\leq\left(\dfrac{2}{\alpha}\varepsilon\right)^{p}$

Putting these two estimates together and noting that ${\varepsilon}$ was arbitrary completes the proof. The proof that ${\underline{D}(\int f,x)=f(x)}$ a.e. is completely analogous. $\Box$

It is easy to see that the above proof breaks down in the endpoint case ${q=1}$. Continuous functions are not dense in ${L^{\infty}}$, so an ${L^{1}}$ estimate for ${\sum\chi_{R_{k}}}$ and Hölder’s inequality is not good enough. If we had some “better” control over the overlap of the ${R_{k}}$, then we would be able to show that the density property holds.

Suppose that instead of the overlap condition for the ${V_{q}}$ covering property stated above, we have condition 3′

$\displaystyle \left\|\sum\chi_{R_{k}}-\chi_{\bigcup R_{k}}\right\|_{L^{q}}<\varepsilon$

Given a measurable set ${A}$ with ${\left|A\right|<\infty}$ and ${0<\alpha<1}$, we will show that

$\displaystyle E_{\alpha,N}:=\left\{x\in A^{c} : \left|x\right|\leq N, \exists\left\{R_{k}(x)\right\}\text{ s.t. }\dfrac{\left|R_{k}(x)\cap A\right|}{\left|R_{k}(x)\right|}>\alpha \ \forall k\right\}$

has measure zero. It is clear that the union of all ${\left\{R_{k}(x)\right\}}$ forms a Vitali cover of ${E_{\alpha,N}}$, whence there exists a countable subcollection satisfying conditions 1,2, and 3′. Observe that

$\displaystyle \begin{array}{lcl} \displaystyle\left|E_{\alpha,N}\right|\leq\sum_{k}\left|R_{k}\right|\leq\alpha^{-1}\int_{A\cap\bigcup R_{k}}\sum\chi_{R_{k}}&=&\displaystyle\alpha^{-1}\int_{A}\left[\sum\chi_{R_{k}}-\chi_{\bigcup R_{k}}\right]+\alpha^{-1}\left|A\cap\bigcup R_{k}\right|\\ [2 em]&\leq&\displaystyle\alpha^{-1}\varepsilon+\alpha^{-1}\left|A\cap\bigcup R_{k}\right|\\ [2 em]&\leq&\displaystyle\alpha^{-1}\varepsilon+\alpha^{-1}\underbrace{\left|\bigcup R_{k}\setminus E_{\alpha,N}\right|}_{\leq\varepsilon}, \end{array}$

since ${A\subset E_{\alpha,N}^{c}}$. Since ${\varepsilon>0}$ was arbitrary, we conclude that ${\left|E_{\alpha,N}\right|=0}$, whence ${E_{\alpha}:=\bigcup_{N}E_{\alpha,N}}$ has measure zero. We conclude that ${D(\chi_{A},x)=0}$ for a.e. ${x\in A^{c}}$. We apply the same argument to the set ${A'=A^{c}}$ and conclude that ${D(\chi_{A^{c}},x)=0}$ for a.e. ${x\in A}$, which implies ${D(\chi_{A},x)=1}$ for a.e. ${x\in A}$.

If we could somehow pass from condition 3 to a countable subcollection satisfying conditions 1,2 and 3′, then the question would be resolved as the argument above shows. But it is not clear to this author how to do that, nor do any potential counterexamples come to mind.