Using Lemma [efr-VH33], there is a unique extension F: \mathsf {Borel}_{\bar {\Delta }} \to \mathcal {C}. There is a faithful (identity on objects) functor \mathsf {Borel}_{\bar {\Delta }} \to \mathsf {BorelStoch} - the only case where this is not full is \operatorname {\mathrm {Hom}}(A,2^\omega ).

It is apparent that we must extend this functor by defining \bar {F}(\phi : A \to 2^\omega ) = \lim _n F(\pi _{2^n}\phi ), using the universal property of F(2^\omega ) = F(2)^\omega . It is apparent that this preserves independent pairings—the only question is whether this extension is actually functorial. By the limit property it suffices to prove that it is functorial for any composable pair A \xrightarrow {\phi } 2^\omega \xrightarrow {\psi } K with K finite.

Let a function f: 2^\omega \to K be good if \bar {F}(\phi ) ; F(f) = F(\phi ; f) for all kernels \phi : A \to 2^\omega (for all A). Let an algebra of sets \mathbb {A} \subseteq \mathcal {B}(2^\omega ) (i.e a collection of subsets stable under finite unions and complements) be called good if every \mathbb {A}-measurable map 2^\omega \to K to a finite set is good. Now we claim:

1. The class \mathbb {A}_0 of sets of the form V \times 2^\omega for V \subseteq 2^N, N finite, is a good algebra.
2. If \mathbb {A} is a good algebra, let \mathbb {A}^+ be the smallest algebra containing all countable unions of sets in \mathbb {A}. Then \mathbb {A}^+ is again a good algebra.
3. Any directed union of good algebras \mathbb {A}_0 \subseteq \mathbb {A}_1 \dots \subseteq A_\alpha \subseteq is again a good algebra.

First note that, by Zorn's lemma, this straightforwardly implies the full Borel \sigma -algebra is a good algebra, which in turn concludes the proof.

Point 3 holds, since both being an algebra and being good are finitary properties (any map to a finite set which is measurable for the union must be measurable at some finite stage). Point 1 is a straightforward consequence of the fact that the projections 2^\omega \to 2^N are good by construction. So we are left with point 2.

Let \mathbb {A} be a good algebra. Let us identify sets by their indicators 2^\omega \to 2. Then we can write any set in \mathbb {A}^+ as g(\vee (f_1^i(-)), \vee f_2^i(-), \dots , \vee f_k^i(-)), where g: 2^k \to 2 is some function, \vee : 2^\omega \to 2 is the indicator of the sequences with at least one 1, and f_j^i, 0 \leq j \leq k, 0 \leq i < \infty are a bunch of \mathbb {A}-measurable functions.

By assumption the pairing of all the f's, 2^\omega \to (2^\omega )^k is good, and so it suffices to show that the mapping (\vee )^k : (2^\omega )^k \to 2^k is good. By induction we may suppose this holds for (\vee )^{k-1}, since it is trivial for k=0.

Now let \phi : A \to (2^\omega )^k be some kernel. Write A = A_0 + A_1, where A_0 is the subset where the probability of only zeroes in the first stream is at least 1/2, and A_1 is the complement, where the probability of at least one 1 in the first stream is >1/2. Observe that by \sigma -continuity, for each a there must be some N so that the probability of at least one 1 in the first N elements of the first stream is at least 1/2. Decompose A_1 into A_1^1 + A_1^2 + \cdots , where for a \in A_1^N there is at least a 1/2 probability of the first 1 being before N. Let \psi _1 : A \to (2^\omega )^k be a kernel which on A_0 is given by (0, \otimes \mu _{2\dots k}) where \mu _{2,\dots k} is the kernel giving the conditional distribution of the remaining streams if the first stream is 0. On A_1^N, it is given by a linear combination \sum _{s \in 2^N \setminus 0} \delta _{s} \otimes \mu _s, where \mu _s is the conditional distribution of the rest of the stream and the remaining streams - we choose this linear combination so that 1/2 of it is \leq the actual probability of each of those prefixes.

Thus we obtain a decomposition of \phi as \sum _i \frac {1}{2^i} \phi _i where for each \phi _i, the distribution on the first coordinate is either concentrated on 0 \in 2^\omega , or concentrated away from 0 on some finite prefix. Each of these satisfy F(\vee ^k)\bar {F}(\phi _i) = F(\vee ^k \phi _i). Hence, since F(\vee ^k) must preserve the infinite linear combination, \vee ^k must be good as desired.

Now let f: 2^\omega \to B be a kernel, and let \phi : A \to 2^\omega again be a kernel. We can factor f as a measurable map \bar {f} : 2^\omega \times 2^\omega \to B, composed with the uniform c^\omega : I \to 2^\omega . By the above we have F(f) = F(\bar {f})\circ (F(c^\omega ) \otimes 2^\omega ). By monoidality, it follows that \bar {F}(f)\bar {F}(\phi ) = F(f\phi ) as desired.