Let $\mu$ be a measure on a measurable space and $K_1, K_2$ be markov kernels satisfying $$ \mu \otimes K_1 (A\times B) = \mu\otimes K_2(B\times A) \qquad \text{for any } A, B \text{ measurable.} $$ I would like to show that this equivalent to the fact that for any measurable functions $f, g$ $$ \mu\otimes K_1(f\times g) = \mu \otimes K_2(g \times f) $$ or in other words $$ \int f(x) g(y) \mu(dx) K_1(x, dy) = \int f(y)g(x) \mu(dy) K_2(y, dx) $$
Attempt
I tried using the standard machinery. First, the assumption is $$ \int_A \mu(dx) K_1(x, B) = \int_B \mu(dy) K_2(y, A) $$ which can also be written in terms of indicator functions $$ \int_{X^2} 1_A(x) 1_B(y) \mu(dx) K_1(x, dy) = \int_{X^2} 1_A(x) 1_B(y) \mu(dy) K_2(y, dx) $$ then my guess is that I could then say that this holds for simple functions and then for measurable functions, by using the Monotone Convergence Theorem. However I have not encountered a problem with this product structure and I am not sure if it would apply.