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Suppose I have a measurable space $(\Omega, \Sigma)$ and a function $f: \mathbb{R} \times \mathbb{\Sigma} \rightarrow [0,1]$ such that for any $x \in \mathbb{R}$ the tuple $(\Omega, \Sigma, f(x, \_))$ is a measure space.

Is it the case that $$\int_{\mathbb{R}} \left[ \int_{\Omega} df(x, \_) \right] dx = \int_{\Omega} d \left[ \int_{\mathbb{R}} f(x, \_) dx \right] \ ?$$

I suspect that this comes down to the question of whether we can interchange the Radon-Nikodym derivative and integration operations, but I am not sure. If this question doesn't make sense because of a missing assumption please let me know :)

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    $\begingroup$ This just might just be my general ignorance, but I'm struggling a bit to make sense of your question. It seems to me like you have a family $(\nu_t)_{t\in \mathbb{R}}$ of measures and want to know whether/when it's possible to make sense of the 'integral' $ \nu = \int_\mathbb{R}\nu_t\hspace{.2pc}\mathrm{d}t$ and, if so, whether $ \nu(E) = \int_\mathbb{R} \nu_t(E)\hspace{.2pc} \mathrm{d}t$ for $E\in \Sigma$. Is that anywhere close? $\endgroup$
    – DCM
    Commented Dec 31, 2021 at 21:06
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    $\begingroup$ (my $\nu_t$ is your $f(t,\cdot)$ by the way - I know I should really stick with your notation, but writing it in a more familiar way might help people (me) recognise certain things more easily) $\endgroup$
    – DCM
    Commented Dec 31, 2021 at 21:07
  • $\begingroup$ @DCM yes I believe that is right $\endgroup$
    – gigalord
    Commented Jan 1, 2022 at 16:16

1 Answer 1

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Provided $x\mapsto f(x,B)$ is Borel measurable for each $B\in\Sigma$, you can define $$ \mu(B):=\int_{\Bbb R} f(x,B) dx, $$ and check (using Tonelli's theorem) that $\mu$ is a measure. If $H\ge 0$ is a measurable function on $\Omega$, then $$ \int_\Omega H d\mu=\int_{\Bbb R}\left[\int_\Omega H(\omega) d_\omega f(x,\omega)\right] dx, $$ by the usual simple function approximation.

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  • $\begingroup$ The OP's notations are somewhat poorly chosen, leading one to commit the error that you yourself make above: the second argument of $f$ is not a point $\omega \in \Omega$, but a measurable subset $B \in \Sigma$, rendering the inner integral on the right side meaningless. $\endgroup$
    – Alex M.
    Commented Jan 1, 2022 at 18:04
  • $\begingroup$ No. My notation is also poorly chosen! The inner integral on the right is intended to be the integral of $H$ with respect to the measure $f(x,\cdot)$. Perhaps I should have used the probabilist's notation $\int_\Omega H(\omega) f(x,d\omega)$. $\endgroup$
    – John Dawkins
    Commented Jan 1, 2022 at 18:15
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    $\begingroup$ Oh, then it's probably $\int _\Omega H \, \mathrm d [f(x, \cdot)]$. $\endgroup$
    – Alex M.
    Commented Jan 1, 2022 at 18:18
  • $\begingroup$ Thank you! I am not sure I follow the last step though. What do you mean by "usual simple function approximation"? $\endgroup$
    – gigalord
    Commented Jan 2, 2022 at 14:06
  • $\begingroup$ The equality is shown for $H=1_B$. By linearity it extends to non-neg simple functions (positive linear combinations of indicators). Arbitrary $H\ge 0$ is the incr. limit of a sequence of simple functions. $\endgroup$
    – John Dawkins
    Commented Jan 2, 2022 at 17:07

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