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My original incorrect claims were too strong, but the argument does give the following, which is perhaps of some interest also:

Suppose that $B$ is closed, and write the complement as a disjoint union of open intervals $B^c=\bigcup I_n$. If $\sum |I_n|\log |I_n|<\infty$, then $\sum a_n(B)<\infty$.

This also says something about arbitrary Borel sets $B$ by using inner regularity.

To prove this, suppose we had an $N$ for which $\sum_{n\le N} a_n$ is very large, and in particular much larger than $\sum |I|\log |I|$. Let $I$ be one of these intervals.

I now want to take a look at what happens to $\sum a_n$ if I add $I$ to $B$. Clearly, your quadratic function of the density is only affected for those dyadic intervals that have a part in $I$ as well as a part in $I^c$. At each level $n$, there are at most two such intervals. If $|I|\ll 2^{-n}$, then the density of a dyadic interval at level $n$ changes by not more than $2^n|I|$. So the change of $a_n$ will be $\lesssim |I|$. There are $\lesssim \log |I|$ such levels $n$. At the others levels $n$, with $|I|\gtrsim 2^{-n}$, it's clear that $a_n$ changes by $\lesssim 2^{-n}$.

We obtain a total change $\lesssim |I|\log |I|$ to $\sum a_n$ when $I$ is included in $B$. By successively adding intervals, we can pass to a new set $C\supseteq B$, of almost full measure, and $$ \sum_{n\le N} a_n(B) \lesssim \sum_{n\le N} a_n(C) + \sum |I|\log |I| . $$ However, $\sum_{n\le N} a_n(C)$ is clearly not very large, so it follows that $\sum_{n\le N} a_n(B)$ can actually not be made arbitrarily large.

My original incorrect claims were too strong, but the argument does give the following, which is perhaps of some interest also:

Suppose that $B$ is closed, and write the complement as a disjoint union of open intervals $B^c=\bigcup I_n$. If $\sum |I_n|\log |I_n|^{-1}<\infty$, then $\sum a_n(B)<\infty$.

This also says something about arbitrary Borel sets $B$ by using inner regularity.

To prove this, suppose we had an $N$ for which $\sum_{n\le N} a_n$ is very large, and in particular much larger than $\sum |I|\log |I|^{-1}$. Let $I$ be one of these intervals.

I now want to take a look at what happens to $\sum a_n$ if I add $I$ to $B$. Clearly, your quadratic function of the density is only affected for those dyadic intervals that have a part in $I$ as well as a part in $I^c$. At each level $n$, there are at most two such intervals. If $|I|\ll 2^{-n}$, then the density of a dyadic interval at level $n$ changes by not more than $2^n|I|$. So the change of $a_n$ will be $\lesssim |I|$. There are $\lesssim -\log |I|$ such levels $n$. At the others levels $n$, with $|I|\gtrsim 2^{-n}$, it's clear that $a_n$ changes by $\lesssim 2^{-n}$.

We obtain a total change $\lesssim -|I|\log |I|$ to $\sum a_n$ when $I$ is included in $B$. By successively adding intervals, we can pass to a new set $C\supseteq B$, of almost full measure, and $$ \sum_{n\le N} a_n(B) \lesssim \sum_{n\le N} a_n(C) + \sum |I|\log |I|^{-1} . $$ However, $\sum_{n\le N} a_n(C)$ is clearly not very large, so it follows that $\sum_{n\le N} a_n(B)$ can actually not be made arbitrarily large.

This is true. Essentially I'm going to claim that if $I=(a,b)\subseteq B^c$ is an interval, then this will be responsible for a contribution of not more than $\lesssim |I|$ to $\sum a_n(B)$, and by approximation, this gives the claim in general.

Here are some additional details. Suppose we had a $B$ such that $\sum_{n\le N} a_n(B)$ becomes arbitrarily large as $N\to\infty$. Fix a large $N$, and approximate $B$ by a compact set $K\subseteq B$ of almost the same measure. Then $\sum_{n\le N} a_n(K)$ is also large. Since $K$ is compact, its complement is a disjoint union of open intervals.

I now want to take a look at what happens to $\sum a_n$ if I add such an interval $I$ to $K$. Clearly, your quadratic function of the density is only affected for those dyadic intervals that have a part in $I$ as well as a part in $I^c$. At each level $n$, there are at most two such intervals, so the change to $a_n$ is $\lesssim 2^{-n}$. In fact, for $I$ to affect densities at level $n$ at all (appreciably), we must have $|I|\gtrsim 2^{-n}$. If $|I|\ll 2^{-n}$, then the density of a dyadic interval at level $n$ could change by $2^n|I|$. So the change in $a_n$ will be $\lesssim |I|$. Since a geometric sum is comparable to its first term, these observations combine to show us that the change to $\sum a_n$ is $\lesssim |I|$ if I include $I$ in $K$.

So we can change $K$ to a set of almost full measure without changing $\sum a_n$ by more than a quantity $\lesssim \sum |I| \le 1$. However, for a set of almost full measure $\sum_{n\le N} a_n$ is clearly not very large, so it follows that it wasn't large to start with.

My original incorrect claims were too strong, but the argument does give the following, which is perhaps of some interest also:

Suppose that $B$ is closed, and write the complement as a disjoint union of open intervals $B^c=\bigcup I_n$. If $\sum |I_n|\log |I_n|<\infty$, then $\sum a_n(B)<\infty$.

This also says something about arbitrary Borel sets $B$ by using inner regularity.

To prove this, suppose we had an $N$ for which $\sum_{n\le N} a_n$ is very large, and in particular much larger than $\sum |I|\log |I|$. Let $I$ be one of these intervals.

I now want to take a look at what happens to $\sum a_n$ if I add $I$ to $B$. Clearly, your quadratic function of the density is only affected for those dyadic intervals that have a part in $I$ as well as a part in $I^c$. At each level $n$, there are at most two such intervals. If $|I|\ll 2^{-n}$, then the density of a dyadic interval at level $n$ changes by not more than $2^n|I|$. So the change of $a_n$ will be $\lesssim |I|$. There are $\lesssim \log |I|$ such levels $n$. At the others levels $n$, with $|I|\gtrsim 2^{-n}$, it's clear that $a_n$ changes by $\lesssim 2^{-n}$.

We obtain a total change $\lesssim |I|\log |I|$ to $\sum a_n$ when $I$ is included in $B$. By successively adding intervals, we can pass to a new set $C\supseteq B$, of almost full measure, and $$ \sum_{n\le N} a_n(B) \lesssim \sum_{n\le N} a_n(C) + \sum |I|\log |I| . $$ However, $\sum_{n\le N} a_n(C)$ is clearly not very large, so it follows that $\sum_{n\le N} a_n(B)$ can actually not be made arbitrarily large.

This is true. Essentially I'm going to claim that if $I=(a,b)\subseteq B^c$ is an interval, then this will be responsible for a contribution of not more than $\lesssim |I|$ to $\sum a_n(B)$, and by approximation, this gives the claim in general.

Here are some additional details. Suppose we had a $B$ such that $\sum_{n\le N} a_n(B)$ becomes arbitrarily large as $N\to\infty$. Fix a large $N$, and approximate $B$ by a compact set $K\subseteq B$ of almost the same measure. Then $\sum_{n\le N} a_n(K)$ is also large. Since $K$ is compact, its complement is a disjoint union of open intervals.

I now want to take a look at what happens to $\sum a_n$ if I add such an interval $I$ to $K$. Clearly, your quadratic function of the density is only affected for those dyadic intervals that have a part in $I$ as well as a part in $I^c$. At each level $n$, there are at most two such intervals, so the change to $a_n$ is $\lesssim 2^{-n}$. Moreover, for $I$ to affect densities at level $n$ at all (appreciably), we must have $|I|\gtrsim 2^{-n}$. A geometric sum is comparable to its first term, so the upshot of all this is that the change to $\sum a_n$ is $\lesssim |I|$ if I include $I$ in $K$.

So we can change $K$ to a set of almost full measure without changing $\sum a_n$ by more than a quantity $\lesssim \sum |I| \le 1$. However, for a set of almost full measure $\sum_{n\le N} a_n$ is clearly not very large, so it follows that it wasn't large to start with.

This is true. Essentially I'm going to claim that if $I=(a,b)\subseteq B^c$ is an interval, then this will be responsible for a contribution of not more than $\lesssim |I|$ to $\sum a_n(B)$, and by approximation, this gives the claim in general.

Here are some additional details. Suppose we had a $B$ such that $\sum_{n\le N} a_n(B)$ becomes arbitrarily large as $N\to\infty$. Fix a large $N$, and approximate $B$ by a compact set $K\subseteq B$ of almost the same measure. Then $\sum_{n\le N} a_n(K)$ is also large. Since $K$ is compact, its complement is a disjoint union of open intervals.

I now want to take a look at what happens to $\sum a_n$ if I add such an interval $I$ to $K$. Clearly, your quadratic function of the density is only affected for those dyadic intervals that have a part in $I$ as well as a part in $I^c$. At each level $n$, there are at most two such intervals, so the change to $a_n$ is $\lesssim 2^{-n}$. In fact, for $I$ to affect densities at level $n$ at all (appreciably), we must have $|I|\gtrsim 2^{-n}$. If $|I|\ll 2^{-n}$, then the density of a dyadic interval at level $n$ could change by $2^n|I|$. So the change in $a_n$ will be $\lesssim |I|$. Since a geometric sum is comparable to its first term, these observations combine to show us that the change to $\sum a_n$ is $\lesssim |I|$ if I include $I$ in $K$.

So we can change $K$ to a set of almost full measure without changing $\sum a_n$ by more than a quantity $\lesssim \sum |I| \le 1$. However, for a set of almost full measure $\sum_{n\le N} a_n$ is clearly not very large, so it follows that it wasn't large to start with.