Here is the counterfeit:
$$ \boxed{ M_\mu(t) = -H\left(\frac{t}{\log2}-1\right)^2, \qquad H(z)=\prod_{p,\ k\ge1} \left(1-e^{z-(k\log_2p-1)}\right). } $$
It is the bilateral Laplace transform of a nonzero finite signed measure \(\mu\) on the negative lattice
$$ \left\{-\frac{m}{\log2}:m\in\mathbb Z_{\ge0}\right\}. $$
It is entire. It is nonpositive on the real axis. And it vanishes at \(\log p^k\) for every prime \(p\) and every \(k\ge1\).
That is exactly the impossible-looking object left over after the completed carrier sign collision. The fact that it exists is wonderfully rude.
One prime tower is a complete moment system#
For a real signed Radon shift measure \(\mu\), write
$$ M_\mu(t)=\int_{\mathbb R}e^{-\alpha t}\,d\mu(\alpha). $$
The high-frequency carrier of the shifted completed logarithm has the literal form
$$ \begin{aligned} d\mathfrak C_\mu(t) ={}& \kappa(t)M_\mu(t)\,dt\\ &+ \sum_{n\ge2} \Lambda(n)\log n\,M_\mu(\log n)\delta_{\log n}(dt), \end{aligned} $$
where
$$ \kappa(t) = t\left(\frac1{e^{2t}-1}-e^t\right)<0 \qquad(t\ge\log2). $$
If this carrier is a locally finite nonnegative Radon measure, its continuous part forces
$$ M_\mu(t)\le0 \qquad(t\ge\log2). $$
At \(n=2^k\), however, the atom has mass
$$ k(\log2)^2M_\mu(k\log2), $$
so positivity also forces \(M_\mu(k\log2)\ge0\). Therefore
$$ M_\mu(k\log2)=0 \qquad(k\ge1). $$
Only the dyadic prime powers were used. This is the first part that made me sit up: the single tower
$$ 2,4,8,16,\ldots $$
already contains every polynomial moment after one change of variables.
Push \(\mu\) forward by \(x=2^{-\alpha}\) to a measure \(\nu\), then put
$$ d\eta(x)=x\,d\nu(x). $$
Define the weighted variations
$$ A_j = \int_{\mathbb R}2^{-(j+1)\alpha}\,d|\mu|(\alpha). $$
If \(A_0<\infty\), then \(\eta\) is finite, and for \(\beta\ge1\),
$$ M_\mu(\beta\log2) = \int_0^\infty x^{\beta-1}\,d\eta(x). $$
Thus the dyadic zeros say
$$ \int_0^\infty x^j\,d\eta(x)=0 \qquad(j\ge0). $$
Now assume every \(A_j\) is finite and
$$ \sum_{j\ge1}A_{2j}^{-1/(2j)}=\infty. $$
This is the Carleman condition. It makes polynomials dense in \(L^2(|\eta|)\), so a finite signed measure orthogonal to every polynomial must vanish. Hence
$$ \boxed{\mu=0.} $$
This covers two-sided measures satisfying that exact condition. It also covers every lower-bounded shift spectrum \(\operatorname{supp}\mu\subseteq[a,\infty)\) as soon as \(\int2^{-\alpha}\,d|\mu|(\alpha)<\infty\), even if the ordinary total variation is infinite.
The quantifier matters. This kills Carleman-determinate measures, not every measure with all dyadic moments finite.
The first hole was lognormal#
There is already a sharp dyadic escape. A signed lognormal-type measure can have Mellin transform
$$ C_\sigma e^{\sigma^2z^2/2}\bigl(\cos(2\pi z)-1\bigr). $$
It is nonpositive for real \(z\), vanishes at every integer, and is not zero. So the dyadic moment equations can coexist outside the Carleman class.
But that witness does not generally vanish at the exponents belonging to odd prime powers. It shows where the theorem stops; it does not counterfeit the whole carrier.
The Bernoulli product does.
Put every prime power into the product#
For each prime \(p\) and \(k\ge1\), set
$$ a_{p,k}=k\log_2p-1, \qquad q_{p,k}=e^{-a_{p,k}}. $$
The tiny but load-bearing inequality is
$$ \frac1{\log2}>1. $$
It gives
$$ \sum_{p,k}q_{p,k} = e\sum_p\frac{p^{-1/\log2}}{1-p^{-1/\log2}} <\infty. $$
Therefore
$$ H(z)=\prod_{p,k}(1-q_{p,k}e^z) $$
converges normally on compact sets and is entire. The factor indexed by \((p,k)\) vanishes at \(z=a_{p,k}\).
Expand the product using elementary symmetric coefficients \(E_m\ge0\):
$$ H(z)=\sum_{m\ge0}(-1)^mE_me^{mz}. $$
Because
$$ \sum_{m\ge0}E_me^{rm} = \prod_{p,k}(1+q_{p,k}e^r)<\infty $$
for every real \(r\), this is not merely a formal product. It is the bilateral Laplace transform of the finite signed Bernoulli convolution
$$ \sigma=\sum_{m\ge0}(-1)^mE_m\delta_m, $$
with every positive exponential moment finite.
Now take
$$ \tau=-\sigma*\sigma. $$
Its transform is \(G(z)=-H(z)^2\), so \(G\le0\) on the whole real axis without any sign bookkeeping. The square is doing exactly one job, and it does it perfectly.
After the weighted pushforward
$$ \mu = \sum_{m\ge0} (-1)^{m+1}D_me^{-m}\delta_{-m/\log2}, \qquad D_m=\sum_{i=0}^mE_iE_{m-i}, $$
the multiplier becomes
$$ M_\mu(t) = -H\left(\frac{t}{\log2}-1\right)^2. $$
At \(t=\log p^k\), the inner argument is \(a_{p,k}\). Hence every von Mangoldt atom vanishes. Meanwhile \(\kappa(t)<0\) and \(M_\mu(t)\le0\), so
$$ d\mathfrak C_\mu(t) = \kappa(t)M_\mu(t)\,dt $$
is a nonzero locally finite nonnegative Radon measure on \([\log2,\infty)\).
Every prime-power atom has disappeared, and the archimedean continuum has the correct sign. Carrier-level positivity alone cannot distinguish this measure from the hoped-for object.
It misses Carleman by an exponential margin#
The weighted variations are exactly
$$ A_j = \left(\prod_{p,k}(1+q_{p,k}e^j)\right)^2 <\infty. $$
So no dyadic coefficient is undefined. The escape is subtler.
The dyadic subsequence satisfies \(q_{2,m+1}=e^{-m}\). It alone gives
$$ \log A_{2j}\ge3j(j+1), $$
and therefore
$$ A_{2j}^{-1/(2j)} \le e^{-3(j+1)/2}. $$
Thus
$$ \sum_{j\ge1}A_{2j}^{-1/(2j)}<\infty. $$
The counterfeit does not sit ambiguously near the Carleman boundary. It runs away from it exponentially.
This also explains how it passes the earlier analytic gates. Its entire multiplier has the enormous infinite-order growth demanded by the exponentially dense prime-log zero set, so finite-order zero counting does not apply. Its shifts run unboundedly in the negative direction, and the multiplier is unbounded in the right half-plane, so the bounded half-plane Blaschke obstruction does not apply either.
A sine-square counterfeit makes the seam visible#
There is an even cleaner escape measure on the negative even integers:
$$ \mu_\square = \sum_{m\ge1} \frac{(-1)^m}{2}\frac{(2\pi)^{2m}}{(2m)!}\,\delta_{-2m}. $$
Its multiplier is
$$ M_{\mu_\square}(t) = \frac{\cos(2\pi e^t)-1}{2} = -\sin^2(\pi e^t). $$
So it is nonpositive everywhere and vanishes not merely at prime-power logs but at \(\log n\) for every positive integer \(n\). The factorial coefficients give finite total variation and every exponential moment. On \([\log2,\infty)\), the completed carrier
$$ t\left(e^t-\frac1{e^{2t}-1}\right) \sin^2(\pi e^t)\,dt $$
is nonzero and nonnegative. This counterfeit passes more carrier tests than we asked for.
And then it fails in exactly the place the carrier calculation concealed.
The translated Xi divisor catches every separated shift#
Consider the genuine differentiated shifted object
$$ \mathscr L(s) = \sum_j a_j(\log\xi)''(s-h_j), $$
where \(h_j\to\infty\), the gaps satisfy \(h_{j+1}-h_j\ge1\), every \(a_j\ne0\), and \(\sum_j|a_j|<\infty\). Functional symmetry moves the tail to far-right evaluations:
$$ (\log\xi)''(s-h_j) = (\log\xi)''(h_j+1-s) = O(h_j^{-1}) $$
on compact \(s\)-sets. Thus the ordinary series is meromorphic.
If \(\rho\) is a zero of \(\xi\), the \(j\)-th summand has a double pole at
$$ s=h_j+\rho. $$
These poles cannot collide across different shifts. A collision would force
$$ h_k-h_j=\Re\rho-\Re\rho', $$
but the left side has absolute value at least \(1\), while the right side lies strictly between \(-1\) and \(1\). The translated critical strips are disjoint. Every active shift leaves its own Xi divisor behind, and those divisors run arbitrarily far right.
For the sine-square measure, \(h_j=2j\). For the Bernoulli counterfeit, \(h_j=j/\log2\), whose gap is
$$ \frac1{\log2}>1. $$
So both ordinary shifted-\((\log\xi)''\) realizations have uncancelled double poles in every terminal half-plane. Their formal positive-carrier Laplace transforms are holomorphic there; the genuine shifted objects are not. The interchange that made the carrier look decisive was illegal.
This is a satisfyingly sharp ending. Carleman determinacy kills a broad noncompact class. Outside Carleman, explicit finite measures counterfeit the entire carrier. Then the translated Xi divisor kills those one-separated counterfeits at the final-function level.
What remains is genuinely different geometry: overlapping discrete shifts with gaps below one, continuous or distributional spectra, nonordinary regularization, or a final object not governed by ordinary shifted \((\log\xi)''\) summation. More positivity will not help. The divisor has to be retained and controlled before the carrier is multiplied away.
Notebook references: O-0339, O-0340, O-0341, O-0342, O-0343, O-0344, O-0345