Conditioning independent variables leaves the marginal law

For each y the section $K(y,\cdot )=P_X$ is a probability measure; for each A the evaluation is constant and therefore measurable. For $B\in \mathcal{T}$, independence through [F3] gives ${\int }_{\{Y\in B\}}K(Y,A)dP=P_X(A)P_Y(B)=P(X\in A,Y\in B)\mathrm{.}$ Every event in $\sigma (Y)$ is of this form, because the inverse images of all measurable B already form a sigma-algebra. This proves [F1], while the first two observations prove [F2].

For a concrete calculation, put probability 1/6 at each point of $\{0,1\}\times \{0,1,2\}$ and let X,Y be the two coordinates. Each pair has probability $(1\mathrm{/}2)(1\mathrm{/}3)$, so their marginal rectangle probabilities factor and the coordinates are independent. The kernel gives $K(y,\{1\})=1\mathrm{/}2$ for every y. For $B=\{0,2\}$ its event integral is $(1\mathrm{/}2)(2\mathrm{/}3)=1\mathrm{/}3$, equal to the mass of the two points $(1,0),(1,2)$. If either variable is deterministic the same formula applies, with a Dirac marginal where appropriate.