Suppose $(X, \Sigma\_X)$ and $(Y, \Sigma\_Y)$ are measurable spaces such that $\Sigma\_Y$ is generated by a set $B$. 


Suppose $k : X \times \Sigma\_Y \to [0, 1]$ has the property that $k(x, -)$ is a (sub-)probability measure for each $x \in X$, and $k(-, E)$ is measurable for every $E \in B$, wrt the Borel $\sigma$-algebra on $[0, 1]$. 


Does that make $k(-, U)$ measurable for each $U \in \Sigma\_Y$?
(in other words, is $k$ a Markov kernel?)



Measurability of kernel on generating set
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The answer is no in general. Indeed, suppose $B=\{E,F\}$ and $k(x,A)$ equals $1/2-f(x),f(x),1/2-f(x),f(x)$ if $A$ is $E\setminus F,E\cap F,F\setminus E,Y\setminus E\setminus F$, respectively, where $f\colon X\to[0,1/2]$ is not $\Si\_X$-measurable. Then $k(\cdot, U)$ is measurable for each $U\in B$ but not for $U=E\cap F\in \Sigma\_Y$. 


However, your desired conclusion will hold if $B$ is an algebra. Indeed, suppose that $B$ is an algebra and consider the set $C$ of all $U\in\Si\_Y$ such that $k(\cdot, U)$ is measurable. Clearly, $C$ is a monotone class. Hence, your 
desired conclusion follows by the [monotone class theorem](https://en.wikipedia.org/wiki/Monotone_class_theorem). 


