Protected set distortion

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A finite trace-to-measure inequality

There are absolute constants $C,Q_0>0$ with the following property. Let $Q\ge Q_0$ be an integer, and install one forbidden residue at each modulus in an arbitrary set $D\subseteq[1,Q]$ of distinct moduli. Let $S$ be the full periodic survivor set and

$$\delta=\mu(S),\qquad \alpha=|S\cap I|/Q,$$

where $I$ is any interval of $Q$ consecutive integers. Set

$$\varepsilon(Q)=C\frac{(\log\log Q)^3}{\log Q}.$$

If $\alpha>\varepsilon(Q)$, then

$$\delta\ge(\alpha-\varepsilon(Q)) \exp\!\left\{-\frac{C\log Q}{\alpha-\varepsilon(Q)}\right\}. \tag{1}$$

In particular, if $\alpha\ge2\varepsilon(Q)$, then

$$\delta\ge\exp(-C'\log Q/\alpha). \tag{2}$$

These are finite density bounds, not logarithmic-density existence theorems. On $I=[Q,2Q)$, all installed congruences are active under the original rule, so no activation discrepancy is hidden in (1). If modulus 1 is installed, then $S$ is empty and the hypotheses requiring positive $\alpha$ are false; hence it may be omitted below.

Checked distortion inputs

The BBMST digest was read first. The following statements and their proofs were checked in the full source: Section 2.2, Lemmas 2.1, 3.3, 3.5, Theorem 3.2, and the Jensen step in the proof of Theorem 3.1. We use their prime-stage distorted measures with parameters $d_p\in[0,1/2]$, denoting the final measure by $\mathbb P$. Define

$$\nu(m)=\prod_{p\mid m}(1-d_p)^{-1},\qquad \Delta=\log_+(d\mathbb P/d\mu).$$

The needed conclusions are:

• Later prime stages preserve all previously exposed coordinate marginals. Setting a stage parameter to zero leaves its measure unchanged.
• The final probability of the forbidden set introduced at prime $p$ is at most $M_p^{(2)}/(4d_p(1-d_p))$, where

$$M_p^{(2)}\le\frac1{(p-1)^2} \prod_{\ell<p}\left(1+ \frac{3\ell-1}{(1-d_\ell)(\ell-1)^2}\right). \tag{3}$$

Distinctness of the original moduli is used in this estimate.

• The mean positive log distortion satisfies

$$\mathbb E_{\mathbb P}\Delta \le2\sum_{q\in D}\frac{\nu(q)}q. \tag{4}$$

If an event $R$ has final probability $b>0$, Jensen's inequality gives

$$\mu(R)\ge b\exp\{-\mathbb E_{\mathbb P}\Delta/b\}. \tag{5}$$

We will preserve the actual union of the small-prime forbidden sets, rather than bound it by their reciprocal sum.

An adjustable prime cutoff

Fix a sufficiently large $y$, and choose

$$d_p=0\quad(p\le y),\qquad d_p=\frac{\log y}{2\log p}\quad(p>y). \tag{6}$$

These are allowed parameters. Elementary prime counting and partial summation show, uniformly in $y$,

$$\sum_{p>y}\frac{d_p}{p} =\frac{\log y}{2}\sum_{p>y}\frac1{p\log p}=O(1). \tag{7}$$

Thus $\nu(p)-1\le2d_p$ and $\prod_p(1+(\nu(p)-1)/p)=O(1)$, uniformly in $y$. If $g$ is the nonnegative squarefree-supported multiplicative function with $g(p)=\nu(p)-1$, then $\nu=1*g$. It follows that

$$\sum_{q\le Q}\frac{\nu(q)}q =\sum_{m\le Q}\frac{g(m)}mH_{\lfloor Q/m\rfloor} \ll H_Q. \tag{8}$$

The logarithm of the product in (3) is at most

$$3\sum_{\ell<p}\frac1\ell+ O\!\left(\sum_\ell\ell^{-2}+\sum_{\ell<p}\frac{d_\ell}{\ell}\right) \le3\log\log p+O(1).$$

Here we used the elementary Euler-product estimates already recorded in coprime entropy. Consequently $M_p^{(2)}\ll(\log p)^3/p^2$. The total final forbidden probability at primes greater than $y$ is therefore at most

$$\eta_y\ll\frac1{\log y} \sum_{p>y}\frac{(\log p)^4}{p^2} \ll\frac{(\log y)^2}{y}. \tag{9}$$

The last step is partial summation with $\pi(t)\ll t/\log t$.

Let $S_y$ be the survivor set using the installed moduli all of whose prime factors are at most $y$, and put $\sigma_y=\mu(S_y)$. Equations (6) and marginal preservation give

$$\mathbb P(S_y)=\sigma_y,\qquad \mathbb P(S)\ge\sigma_y-\eta_y.$$

Combining (4), (5), and (8), for $t=\sigma_y-C_1(\log y)^2/y>0$, yields

$$\delta\ge t\exp(-C_2\log Q/t). \tag{10}$$

The function $t\mapsto te^{-K/t}$ is increasing, which justifies using a lower bound for the final surviving probability.

Reading the small-prime measure from the integer interval

Put

$$M=\exp\!\left(\frac{\log Q}{4\pi(y)}\right).$$

The subsystem with $y$-smooth moduli at most $M$ has a period $P$ satisfying

$$P\mid\prod_{p\le y}p^{\lfloor\log_pM\rfloor},\qquad \log P\le\pi(y)\log M=\tfrac14\log Q.$$

Its survivor set $S_{y,M}$ contains $S$. Periodic counting on the interval $I$, uniformly in its starting point, gives

$$\mu(S_{y,M})\ge\alpha-Q^{-3/4}. \tag{11}$$

The reciprocal sum of all $y$-smooth integers greater than $M$ obeys

$$\sum_{\substack{m>M\\P^+(m)\le y}}\frac1m \ll (\log y)\exp(-\log M/\log y). \tag{12}$$

Indeed, Rankin's inequality with $\theta=1/\log y$ gives the upper bound $M^{-\theta}\prod_{p\le y}(1-p^{-1+\theta})^{-1}$. For $y$ large enough that $\theta\le1/4$, comparison with $\prod_{p\le y}(1-1/p)^{-1}$ shows this product is $O(\log y)$: the first-power logarithmic difference is bounded by $e\theta\sum_{p\le y}(\log p)/p=O(1)$, and the higher powers are bounded by an absolutely convergent series.

The union bound for the omitted smooth moduli and (11)--(12) imply

$$\sigma_y\ge\alpha-Q^{-3/4} -C_3(\log y)\exp\!\left\{-\frac{\log Q}{4\pi(y)\log y}\right\}. \tag{13}$$

Finally set $L=\log\log Q$ and $y=\log Q/(A_0L)$, with a large absolute $A_0$. Chebyshev's bound $\pi(y)\le C_\pi y/\log y$ makes the exponent in (13) at most $-4L$ when $A_0\ge16C_\pi$. Its error is then $Q^{-3/4}+O(L/(\log Q)^4)$. Meanwhile (9) is $O(L^3/\log Q)$. Substitute into (10) to prove (1). For (2), use $\alpha-\varepsilon\ge\alpha/2$ and absorb the prefactor by $\alpha\log(2/\alpha)=O(1)$.

A smaller error using quasipolynomial equidistribution

Combining the preceding distortion argument with the new interval equidistribution theorem improves the error in (1) to

$$\varepsilon(Q)=\exp\{-c(\log Q)^{1/6}\} \tag{14}$$

for an absolute $c>0$ and sufficiently large $Q$, leaving the constant $C$ in the exponent of (1) absolute. In particular (2) holds whenever $\alpha\ge2\varepsilon(Q)$ with this smaller error. This improvement uses Bazzi's theorem through the linked interval result; the earlier logarithmic error has the elementary smooth-period proof above.

Let $A$ be the interval theorem's constant and choose

$$v=\left(\frac{\log Q}{2A}\right)^{1/3},\qquad M=e^v,\qquad y=e^{\sqrt v}.$$

The interval of length $Q$ is long enough for that theorem with largest modulus $M$, since $\exp(A(\log M)^3)=Q^{1/2}$. Thus (11) is replaced by $\mu(S_{y,M})\ge\alpha-O(e^{-2v})$. The same smooth-tail estimate (12) and distortion loss (9) give, respectively,

$$\sigma_y\ge\alpha-O(e^{-2v}+\sqrt v\,e^{-\sqrt v}), \qquad \eta_y=O(v e^{-\sqrt v}).$$

For large $Q$ their total error is at most (14), after reducing $c$. Equation (10) proves the improved assertion.

Limitation

For fixed positive $\alpha$, (2) allows the full survivor density to go to zero as a power of the largest modulus. It neither bounds the incremental Haar mass of arbitrary first-kill cells in terms of their logarithmic trace nor proves the required tail continuity. The sparse range below (14) is not treated by these estimates.