We proceed by induction on k, the case k=0 being trivial.

First, consider a kernel \alpha : A \to (2^\omega )^k which, on the first N bits in the first coordinate, is concentrated on a given string a = a_1,\dots a_N which contains at least one 1. This kernel factors over the inclusion of this component: (2^\omega )^k = \{a\} \otimes 2^{\{N+1, \dots \}} \otimes (2^\omega )^{k-1} \sqcup (2^N \setminus \{a\}) \otimes 2^{\{N+1, \dots \}} \otimes (2^\omega )^{k-1}

This factorization is preserved by F (which preserves coproducts). Note that \vee ^k, when restricted to this summand, factors over the projection (2^\omega )^k \to (2^\omega )^{k-1}. So we can write F(\vee ^k)\bar {F}(\alpha ) = F((1,\vee ^{k-1}) \pi )\bar {F}(\alpha ). It is clear that F(\pi )\bar {F}(\alpha ) = \bar {F}(\pi \alpha ), and so by induction this means F(\vee ^k)\bar {F}(\alpha ) = F(\vee ^k \alpha ).

By an analogous argument (without passing through the finite truncation), if \alpha is concentrated on the string 0 \in 2^\omega in the first coordinate, again F(\vee ^k)\bar {F}(\alpha ) = F(\vee ^k \alpha )

Now consider the generic kernel \phi of the lemma. Let A = A_0 \sqcup A_1, where A_0 consists of those a for which the probability that the first coordinate is zero is \geq 1/2, and A_1 is the complement, consisting of those for which at least one 1 has probability >1/2. Note that by \sigma -continuity, there exists for each a \in A_1 some N_a so that there is a probability \geq 1/2 of an 1 in the first N_a coordinates. Note that the least such N_a is a measurable function of a, and so A_1 decomposes into a sum A_1 = A_1^1 \sqcup A_1^2 \dots , where on A_1^N there is a \geq 1/2 probability of an 1 in the first N coordinates.

Now let \mu _1 be the kernel defined as follows:

1. On A_0, it is deterministically 0 in the first coordinate, and given by the conditional distribution given this in the last k coordinates.
2. For a \in A_1^N, choose some distribution \nu on 2^N so that \nu / 2 is less than the marginals of \phi , and so that \nu is concentrated away from 0 (this is possible by construction of A_1^N). Then for each b \in 2^N, let \xi _b be the measure which is concentrated on that bitstring on the first N bits, and given by the conditional distribution according to \phi everywhere else. Let \mu _1(a) = \sum _{b\in 2^N} \nu (b) \xi _b

Observe that by construction, \mu _{1/2} \leq \phi . Now we can continue the recursion, defining \mu _2, \mu _3, \dots , and we must necessarily have \sum _i 1/2^i \mu _i = \phi . This sum must be preserved by \bar {F} (where the meaning of this sum in \mathcal {C} is given by the canonical sampling morphism X^\omega \to X, or equivalently by the dual I \to \omega ). Therefore we have F(\vee ^k)\bar {F}(\phi ) = F(\vee ^k)(\sum _i 1/2^i \bar {F}(\mu _i)) = \sum _i 1/2^i F(\vee ^k) \bar {F}(\mu _i) = \sum _i 1/2^i F(\vee ^k \mu _i) = F(\vee ^k \phi ), finishing the proof.