Suppose $\mu$ is a probability measure and $K$ a Markov kernel, on the same space. I feel like the following statement is true: $$ \forall\, \text{measurable } A\,:\, \quad \quad \text{if }\left(\int K(x, A) \mu(dx) = 0\right)\implies \left(K(\cdot, A) = 0 \text{ for } \mu\text{-almost every }x\right) $$ Notice that $A$ is fixed and I am saying that whenver the LHS holds, then must the RHS. I could not prove it formally, however.
Attempted Proof
I'll call the measurable space $(X, \mathcal{X})$. The statement on the RHS means $$ \mu\left(\left\{x\in X\,:\, K(x, A) = 0\right\}\right) = 1. $$ I can write this statement as follows $$ \int_X \mathbf{1}_{\{0\}}(K(x, A)) \mu(dx) = 1 \qquad \text{ or } \qquad \int_X \mathbf{1}_{(0,1]}(K(x, A)) \mu(dx) = 0 $$ But I am unsure how to get either of these from the LHS.
This comes from the more general result in measure theory, if $f\geq 0$ and $\int_X fd\mu = 0$ then for $\mu(dx)$ almost every $x\in X$ we have $f(x)=0$.
Proof: Use Markov inequality :
For all $n\in \mathbb{N}^*$ we have $$\mu(\{x\in X \mid f(x)\geq \frac{1}{n} \}) \leq n\int_X f(x)\mu(dx) = 0$$
Doing the countable union over all $n\geq 1$ we obtain that $$\mu(\{x \in X \mid f(x) > 0\}) = 0$$ as wished.
Apply this to $f = K(\;.\;, A)$