Integration with respect to the product of a probability measure and a Markov kernel

Let

• $$(\Omega_i,\mathcal A_i)$$ be a probability space
• $$\mu$$ be a probability measure on $$(\Omega_1,\mathcal A_1)$$
• $$\kappa$$ be a Markov kernel with source $$(\Omega_1,\mathcal A_1)$$ and target $$(\Omega_2,\mathcal A_2)$$

Note that $$(\mu\kappa)(A_2):=\int\mu({ d}\omega_1)\kappa(\omega_1,A_2)\;\;\;\text{for }A_2\in\mathcal A_2$$ is a probability measure on $$(\Omega,\mathcal A_2)$$.

I've read that, by the Cauchy-Schwarz inequality, $$\int\mu({\rm d}\omega_1)\int\kappa(\omega_1,{\rm d}\omega_2)f(\omega_2)\le\int f\:{d}(\mu\kappa)\tag1$$ for all $$\mathcal A_2$$-measurable $$f:\Omega_2\to[0,\infty)$$. However, it's obvious that we've got equality in $$(1)$$ for any elementary $$\mathcal A_2$$-measurable $$f$$, so shouldn't we trivial obtain equality for all $$\mathcal A_2$$-measurable $$f$$? What am I missing? I don't see how the Cauchy-Schwarz inequality is needed here.

• You are absolutely right. What you have read makes no sense at all. There is no need for C-S and the inequality is actually an equality for any non-negative measurable function $f$. – Kabo Murphy Oct 26 '18 at 23:30