Set *d*(*x*)=*f*(*x*)-*f*(*x*+1), with
*f*(1) taken to be 1. The sum over all *x* is 1, because
*f*(1)=1 and *f* tends to zero. Each *d*(*x*) is positive
by the assumption that *f* is monotone strictly decreasing, so
*d* satisfies the condition of full support.

The value of the sum of *d*(*k**A*) cannot be greater than the
sum of *d*(*x*) over all *x*≥*A*, so
1-*d*(1)-*d*(2)-...-*d*(*A*-1)=1-(*f*(1)-*f*(2))-(*f*(2)-*f*(3))-...-(*f*(A-1)-*f*(*A*))=*f*(*A*).

Q.E.D.

In defining *d*(*x*), we'll consider two cases.

In the first case, there exists a *k*>0 such that *k*!=*x*.
In this case, we'll define
*d*(*x*)=2^{-x}(1-*g*(1))+*g*(*k*)-*g*(*k*+1).

In all other cases, we'll define
*d*(*x*)=2^{-x}(1-*g*(1)).

First, let us show that this is a distribution. To prove this, consider that
the sum of all 2^{-x}(1-*g*(1)) equals 1-*g*(1)
while the sum of all *g*(*k*)-*g*(*k*+1) is
*g*(1), so together they sum up to 1. Second, both
2^{-x}(1-*g*(1)) and *g*(*k*)-*g*(*k*+1)
are always positive by the fact that *g*(1) is smaller than 1 and
*g* is strictly monotone decreasing. So, this is a distribution with
full support, as desired.

To see that this distribution meets our criteria, consider only the
*g*(*k*)-*g*(*k*+1) part. For any *A*, all
*k*! values with *k*≥*A* divide by *A*, so
*p*(*A*) is at least the sum of all
*g*(*k*)-*g*(*k*+1), which is to say at least
*g*(*A*), as desired.

Q.E.D.

Let us name the prime numbers sequence: *P*_{0}=2,
*P*_{1}=3, *P*_{2}=5, etc..

Consider, now, *p*(*A*)=1/*A* for some constant *A*.
This is a sum of *d*(*x*) over all *x* that are
multiples of *A*.

Let us calculate how much the sum of *d*(*x*) is over
*x* values that divide by *A* but not by
*P*_{0}*A*,...,*P*_{k}*A*
for some *k*.

If *k*=0, we can do this simply by *p*(*A*)-*p*(2*A*).
However, if
we want to increase *k* to 1, simply subtracting *p*(3*A*) will
cause the contribution due to values dividing by 6*A* to be discounted
twice. In order to calculate this properly, we will need to add this value back.

In general, the result is the inclusion-exclusion formula: the total
probability assigned to multiples of *A* not dividing any of
*P*_{0}*A*,...,*P*_{k}*A* equals

*p*(*A*)-Σ_{i=0}^{k}
*p*(*P*_{i}*A*)
+ Σ_{i=0}^{k}
Σ_{j=i+1}^{k}
*p*(*P*_{i}*P*_{j}*A*) - ...
=1/*A*-Σ_{i=0}^{k} 1/(*P*_{i}*A*)
+ Σ_{i=0}^{k}
Σ_{j=i+1}^{k}
1/(*P*_{i}*P*_{j}*A*) - ...
=1/*A* Π_{i=0}^{k}
(1-1/*P*_{i})

This product is well known to tend to zero as *k* tends to infinity,
so the probability that *e* equals *A* exactly, *d*(*A*),
must be zero. This not only contradicts our assumption of full support, but
also the assumption that this is a distribution, as its sum over all *A*
values will be zero, not 1.

Q.E.D.