Measure-valued maps equal a.e. if their integrals equal over any set Let $X,Y$ be two standard Borel spaces and let $p,q:X\to\mathcal P(Y)$ be two stochastic kernels, which can be alternatively seen as measure-valued maps. Suppose that for some measure $\mu\in \mathcal P(X)$:
$$
  \int_{A}(p(x) - q(x))\mu(\mathrm dx) = 0 \qquad \forall A\in \mathcal B(X)
$$
that is the integral above is a zero measure in $\mathcal M(Y)$. Does it imply that $p =q$ holds true $\mu$-a.e.?

Here I denote $\mathcal B(X)$ to be the Borel $\sigma$-algebra of $X$ and $\mathcal M_b(X)$ is the space of all bounded signed measures on $X$ endowed with a $\sigma$-algebra generated by evaluation maps. In particular, $\mathcal P(X)\subset \mathcal M_b(X)$ is the space of all probability measures over $X$. 
I guess, the integral above can be considered in general as a Bochner integral and perhaps I'm asking about properties of the latter.
 A: If a vector measure has a Radon-Nikodym derivative, then this derivative is unique (it may not exist in general though). Since the constant function with value $0$ is a Radon-Nikodym derivative of the signed measure $A\mapsto \int_A p-q~d\mu$, we have $p=q$ $\mu$-almost surely, provided $p-q$ is Bochner-integrable, which amounts to being measurable, being the almost surely limit of integrable simple functions and taking almost surely values in a separable subspace. In this case, $\{0\}$ is such a subspace.
One can get the desired result without any vector-measur theory though. There is countable algebra $\mathcal{A}$ that generates $\mathcal{B}(Y)$. For each $A\in\mathcal{A}$, we have $p(x)(A)=q(x)(A)$ for all $x$ outside a $\mu$-null set $N_A$. So on the full $\mu$-measure set $X\backslash \bigcup_{A\in\mathcal{A}}N_A$, $p$ and $q$ agree on all of $\mathcal{A}$ and hence on all of $\mathcal{B}(Y)$.
This is pretty much the standard proof that disintegrations are unique up to null sets.
