By Lemma [efr-X0VA], we see that we can immediately flatten any nested sequence in BT(A)^{\omega ^n} into a sequence of dyadic binary trees in BT(A)^\omega , without altering either the sampling map or the represented distribution in \bar {\Delta (A)}. Hence it suffices to prove that the sampling map BT(A)^\omega \to A factors over \bar {\Delta (A)}.

It is clear that the sum map BT(A)^\omega \to \bar {\Delta }(A) is surjective. We will begin by choosing a particular splitting of this map, assigning to each countably supported measure a "normal form" in BT(A)^\omega

Fix an arbitrary total ordering on A. If A is countable there is no difficulty doing this, if A = 2^\omega , we may use the lexical ordering.

Given a finitary distribution \mu , we compute its normal form as follows:

1. Let N_0 be the smallest natural so that there exists a dyadic distribution \nu _0 = \sum _i \frac {k_i}{2^N} a_i so that \frac {1}{2}\nu _0 < \mu - that is, for every element of A, the probability under \frac {1}{2}\nu _0 is strictly less than the probability under \mu .
2. Let \nu _0 be a dyadic distribution as above, choosing the lowest possible a_i according to the chosen ordering. Represent \nu _0 as a balanced binary tree of minimal depth, with the leaves increasing left to right.
3. Repeat these steps with \mu - \frac {1}{2}\nu _0 to construct \nu _1, \nu _2, \dots .

It is clear that this defines a measurable map \Delta (A) \to BT(A)^\omega , and that this is a section of the map \sum _i 1/2^i \nu _i in the other direction.

1. Let S: BT(A)^\omega \times (2^{\omega } \setminus 0) \times (2^\omega )^\omega \to A be the sampling map, which uses the first bitstream to choose the index i from the sequence, then the ith one of the remaining streams to sample from the given binary tree.
2. Note that by construction, if \nu _1 is the first element in the normal form of \mu , and \lambda _i is another sequence of binary trees so that \sum _i 1/2^i \lambda _i is equal to \mu , then there exists some finite prefix so that 1/2 \nu _0 < \sum _{i=0}^N 1/2^i \lambda _i. Then there exists some rearrangement of this prefix \lambda _i' so that \lambda _1' = \nu _1 (and \lambda '_i = \lambda _i when i>N). Let \mathrm {nm}_k: BT(A)^\omega \to BT(A)^\omega be the map which puts the first k elements into the normal form like this. Note that for each k there exists some permutation \sigma _k : BT(A)^\omega \times (2^{\omega } \setminus 0) \times (2^\omega )^\omega \to (2^{\omega } \setminus 0) \times (2^\omega )^\omega which, for each possible sequence, witnesses this rearrangement (in the same way as Lemma [efr-X0VA]).
3. By construction the nth coordinate of \mathrm {nm}_k(a) is equal for all k > n, and so there is a well-defined limiting map \mathrm {nm}_\infty : BT(A)^\omega \to BT(A)^\omega , which (by construction) carries a sequence to its normal form.
4. For \sigma _k, similarly note that the first n bits (deciding the index into the sequence) are fixed for k > n, and the same is true for the n first streams used to sample from the given binary trees.
5. Let \mathrm {nm}_\infty and \sigma _\infty be the limiting maps.

Now observe that F(a,b) = F(\mathrm {nm}_\infty (a),\sigma _\infty (a,b)), and that \sigma _\infty (a,b) for each a preseves the uniform measure. Hence just as before, this proves that the sampling map commutes with taking the normal form, as we wanted.

Observe that (for any monad), the Kleisli category is presented by adding morphisms s_A: TA \to A for each A, subject to the equations s_A\eta _A = 1_A and s_A s_{TA} = s_A \mu _A. We have just constructed the s_A, so we now have to prove that they satisfy these equations. Unitality is clear. The multiplicativity condition follows by observing that any finitary distribution on \Delta (A) can be decomposed as an iterated sum coming from \Delta (A)^{\omega ^N}, at which point this reduces to the mentioned \omega -midpoint homomorphism property.