Alpha i deletes the i-th smallest element of a finite set of positive integers. Theorem A asks for a finitely supported random set whose law moves by less than epsilon, in total variation, under each of the first k deletions.

The natural candidate is a walk with independent, identically distributed steps. It fails. For any honest, countably additive step law, deleting the first point moves the law by more than one quarter. That proof comes later.

So the proof detours through finitely additive step laws, called means, where the invariance is exact. An average over set sizes, and two tools from functional analysis, then return a finitely supported law.

A mean is a positive linear functional on bounded sequences, sending the constant one to one. On indicators it is a finitely additive probability. Probability vectors are examples.

Take the uniform law on one through n, the Cesaro average, and let n grow. The bars flatten and spread right.

A finite window holds at most its size over n, so it empties. The total stays one.

Means form a weak star compact set, so this sequence has an accumulation point, nu. It vanishes on every finite set, which the paper calls diffuse: a random very large number. A short argument, ours, puts its mass at infinity, in the Stone Check compactification on the board.

Average f and its shift over one through n. Only the two end terms differ, so the gap is at most two over n times the supremum norm of f. Nu is therefore shift invariant.

Condition two i evaluates f at a sum of two draws from nu. For fixed x the inner mean sees a shifted f and returns nu of f. The outer mean averages a constant. Hence nu star nu is nu.

No honest law on the positive integers does this. Two draws sum to at least twice the smallest atom, so the convolution puts nothing there. The paper states that an idempotent mean is diffuse. The same check, ours, proves it.

Let f be one when x is less than y: the region above the diagonal.

Fix x and average over y. Up any column all but finitely many cells are shaded, and a diffuse mean ignores finite sets. The iterated mean is one.

Average over x first. Along any row only finitely many cells are shaded, so the iterated mean is zero. The example is ours. The paper says only that any form of Fubini fails.

So a product of means must fix an order. The paper uses the Arens convolution, with x prime inside and x outside, and warns that the independent walk is no formal definition.

Nu k plus one is defined by recursion. Freeze a last step x, and let f upper x append the point y k plus x. Apply nu k, then average over x.

Unrolled for three points, the last step is outermost and the first step innermost. That nesting is part of the definition.

The innermost mean runs with the outer variables frozen. Our computation: under nu two, the first step exceeds the second with mean probability one. Honest independent steps with one common law give at most one half.

In our reading the steps are ordered by scale: each is very large, dwarfs the next, and alone has law nu.

A deletion acts on a mean by push forward. To test alpha j nu on f, give nu the function f composed with alpha j. Delete first, then evaluate.

Take the last point, j equal to k plus one. Inside the recursion we append the point y k plus x, and alpha deletes that same point. The tuple is unchanged, so f composed with alpha, upper x, equals f. No x is left.

The outer mean averages a constant and returns it. So deleting the last point of nu k plus one gives nu k. Only normalization was used. Idempotence has not entered yet.

Delete point k, the second to last. Expand the recursion twice: x is the last step, x prime the step before.

Alpha k removes the point between them. That leaves y k minus one, then one point at distance x plus x prime. The expanded function is f upper x plus x prime.

The two arcs merge. Let g of z be nu k minus one applied to f upper z. For k equal to one, g is f itself.

We now have nu over x, of nu over x prime, of g at the sum. Condition two i collapses it to nu of g, which is nu k of f. Idempotence enters only here.

For an earlier point, j at most k minus one, peel off the last step. Append the point y k plus x, then delete entry j.

Or delete first, then append. The last entry is still y k, so both routes give the same tuple.

So alpha j moves inside the recursion, onto nu k. Induction turns that into nu k minus one, and the recursion rebuilds nu k of f. Steps one and two cover the top two deletions at every level, and all of level one.

For j equal to k the routes differ, since deleting y k changes which entry is last. That is our reading of why that case is separate.

Theorem two point three relates neighboring levels: a deletion sends nu i to nu i minus one. To get one invariant object, average n consecutive levels, all viewed as means on sets with at least k elements.

Apply alpha j with j at most k. Every box shifts one place to the left.

The middle boxes cancel. What is left is nu k minus nu k plus n, divided by n, with norm at most two over n.

Take an accumulation point, nu bar. A routine check, ours: push forward is weak star continuous, as the adjoint of composition with alpha j. So nu bar is exactly invariant under the first k deletions.

Nu bar is still a mean. Goldstine's theorem: finitely supported probabilities are weak star dense in the means. The paper cites it. This Hahn Banach proof is ours.

Suppose a mean m lies outside the closed convex hull of point masses. Separating functionals are evaluations at a bounded f, so some f gives m of f above the supremum of f.

Positivity forbids that. The supremum of f, minus f, is nonnegative, so m of f is at most that supremum.

Hence a net of finitely supported mu q converges weak star to nu bar. The defects, alpha i mu q minus mu q, lie in little l one and go to zero weakly.

Stack the k defects of mu q into one vector in k copies of little l one. These vectors go to zero weakly. Their norms need not shrink.

Mah zoor's trick: a convex set has the same weak and norm closure. By Hahn Banach, a norm closed convex set is an intersection of half spaces, which are weakly closed.

Defects of convex combinations form a convex set with zero in its weak closure. So some combination, mu tilde, has all k defects below epsilon in norm. The paper cites Day. The details are ours.

Mu tilde is finitely supported, and total variation is half the l one distance. That is Theorem A.

Now the promised lower bound. Let M be the first place where the cumulative step law passes one half. The first point is at most M with probability above one half.

Pull that event back through alpha one: the second point is at most M. Then the first step and the second step are each at most M minus one.

Those events are independent, each with probability at most one half. So the pulled back event has probability at most one quarter, and the law moved by more than that.

Means escape this at a price: no construction, no rate, and by our count at least four uses of choice.