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This builds off on an old question about mixture measures: Generalized notions of mixture

Suppose $\mathcal{M}$ is a family of probability measures, and $Q$ is a probability measure over $\mathcal{M}$. We can define a mixture measure by: $$ \mu(A) = \int_A \nu(A)\,dQ(\nu). $$

Given a measurable function $g$, its integral wrt $\mu$ is naturally given by $\int g\,d\mu$. My question is: Is there a way to write this integral in terms of $\nu$ and $Q$?

I think I am just missing something very simple: For example, suppose each $\nu\in\mathcal{M}$ has a density function $f_\nu$, so that $\mu$ also has a density given by $f_{\mu}=\int f_\nu\,dQ(\nu)$. Then we can write (assuming all operations are justified, etc.): \begin{align*} \mathbb{E}_{\mu} g = \int g\,d\mu &= \int g(z)f_{\mu}(z)\,dz \\ &= \int g(z)\int f_{\nu}(z)\,dQ(\nu)\,dz \\ &= \iint g(z) f_{\nu}(z)\,dz\,dQ(\nu) \\ &= \int \mathbb{E}_{\nu}g(z)\,dQ(\nu). \end{align*}

I'm curious if a similar calculation works without assuming the existence of density functions on $\mathcal{M}$. I don't see why not, but for some reason I can't get this to work.

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  • $\begingroup$ There are formulae of this type. The formula is simply $\int g\,d\mu=\int(\int g\,d\nu)\,d\Q(\nu)$. This automatically holds for simple functions from your definition, so can be extended to positive functions in the usual way etc etc. $\endgroup$
    – Anthony Quas
    Commented Mar 12, 2017 at 19:02
  • $\begingroup$ When you say "there are formulae", do you mean that there are standard references for this type of result? If so, I'd be interested in these. Thanks for the suggestion to use approximate with simple functions, this seems to work. I wonder if there is a more direct proof, though? $\endgroup$
    – JohnA
    Commented Mar 12, 2017 at 21:32
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    $\begingroup$ The context where I know this is called ergodic decomposition. It is not as general as what you are looking for: all of the measures $\nu$ that appear are ergodic invariant measures with respect to some transformation in that setting. Note that to make sense of $\mu$ as a mixture, you need that $\int g(z)\,d\nu(z)$ is a measurable function of $\nu$ for each $g$ in some reasonable class (such as the continuous functions). $\endgroup$
    – Anthony Quas
    Commented Mar 12, 2017 at 22:40
  • $\begingroup$ The formula that Anthony gave in his comment does hold in general, and it follows by recalling how the integral was defined, starting from simple functions and then increasing generality gradually. For simple functions the desired identity holds by fiat, so you obtain it for arbitrary integrable functions by following this procedure. $\endgroup$
    – Christian Remling
    Commented Mar 13, 2017 at 0:41

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