The finite POMDP Bayes filter is a probability kernel

For fixed $s,a$, the mass $J(s^{\mathrm{\prime }},\rho ,o\mid s,a)=p(s^{\mathrm{\prime }},\rho \mid s,a)O(o\mid s^{\mathrm{\prime }})$ is nonnegative. Summing first over observations gives ${\sum }_{s^{\mathrm{\prime }},\rho ,o}J={\sum }_{s^{\mathrm{\prime }},\rho }p(s^{\mathrm{\prime }},\rho \mid s,a){\sum }_{o}O(o\mid s^{\mathrm{\prime }})={\sum }_{s^{\mathrm{\prime }},\rho }p(s^{\mathrm{\prime }},\rho \mid s,a)=1$ by [F1]; all index sets are finite, so the reordering is justified by [F6] and [F7]. Thus the joint next-state/reward/observation mass is normalized.

The predicted coordinates ${\bar{b}}_{b,a}(s^{\mathrm{\prime }})$ are nonnegative, and ${\sum }_{s^{\mathrm{\prime }}}{\bar{b}}_{b,a}(s^{\mathrm{\prime }})={\sum }_{s}b(s){\sum }_{s^{\mathrm{\prime }}}T(s^{\mathrm{\prime }}\mid s,a)={\sum }_{s}b(s)=1$ by [F1], [F2], and finite double-sum interchange [F7]. Hence ${\bar{b}}_{b,a}$ is a probability distribution.

Every $q(o\mid b,a)$ is nonnegative, and ${\sum }_{o}q(o\mid b,a)={\sum }_{s^{\mathrm{\prime }}}{\bar{b}}_{b,a}(s^{\mathrm{\prime }}){\sum }_{o}O(o\mid s^{\mathrm{\prime }})={\sum }_{s^{\mathrm{\prime }}}{\bar{b}}_{b,a}(s^{\mathrm{\prime }})=1$ by [F1], step 1.2, and [F7]. Thus $q(\cdot \mid b,a)$ is a probability distribution, so each $q$ is finite and lies in $[0,1]$.

Fix $o$ with $q(o\mid b,a)>0$. Apply [F4] with prior $\pi ={\bar{b}}_{b,a}$ on the finite state set, observation kernel $Q(s^{\mathrm{\prime }},\cdot )=O(\cdot \mid s^{\mathrm{\prime }})$, counting measure on finite $\mathrm{\Omega }$, and density $\mathrm{\ell }(s^{\mathrm{\prime }},o)=O(o\mid s^{\mathrm{\prime }})$. The density is nonnegative and jointly measurable, and its marginal is exactly $q(o\mid b,a)$, which is finite and positive by step 2.1. The theorem's posterior is therefore $\tau (b,a,o)$, with coordinates summing to one. At $q=0$ the Bayes theorem supplies a filler for its everywhere kernel, but the partial update defined here remains undefined and $K$ omits that zero-mass observation.

For fixed $b,a$, the formula for $K$ is a finite nonnegative sum of point masses at the defined $\tau (b,a,o)$. For pairwise disjoint Borel sets $(C_n{)}_{n\ge 1}$, each point $\tau (b,a,o)$ lies in at most one $C_n$, so $1_{{\cup }_n{C}_n}(\tau )={\sum }_n{1}_{C_n}(\tau )$ with at most one nonzero term. Since there are finitely many observations, distributing the finite weighted sum over these finitely many nonzero terms gives $K({\cup }_n{C}_n)={\sum }_{n}K(C_n)$. Also $K(\mathrm{\Delta }(S))={\sum }_{o:q>0}q(o\mid b,a)=1$ by step 2.1. Thus $K(\cdot \mid b,a)$ is a probability measure.

On $\mathrm{\Delta }(S)\times A$, the predicted law $\bar{K}(\{s^{\mathrm{\prime }}\}\mid b,a)={\bar{b}}_{b,a}(s^{\mathrm{\prime }})$ is a probability kernel: step 1.2 gives probability sections, and each evaluation is a finite linear function of $b$ on each discrete action. The function $((b,a),s^{\mathrm{\prime }})\mapsto O(o\mid s^{\mathrm{\prime }})$ is product-measurable, so [F5] makes $q(o\mid b,a)=\int O(o\mid s^{\mathrm{\prime }})\bar{K}(d{s}^{\mathrm{\prime }}\mid b,a)$ measurable; its displayed finite formula also shows it is continuous in $b$. On the Borel set where $q(o\mid b,a)>0$, every coordinate of $\tau$ is a quotient of continuous functions with positive denominator and is Borel. For Borel $C$, the function equal to $q(o\mid b,a)1_C(\tau (b,a,o))$ there and zero where $q=0$ is Borel; summing these functions over the finite set $\mathrm{\Omega }$ proves that $K(C\mid b,a)$ is Borel measurable. By [F3], $K$ is a probability kernel.

The only conditional law used in step 3.1 is the explicit dominated-kernel formula at a positive finite marginal. The finite history beliefs are ratios on positive-probability atoms, and all other constructions use specified finite sums and kernels; no version selection or arbitrary-family choice is used.