Timeline for Does the Radon-Nikodym derivative commute with integration?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 2, 2022 at 17:07 | comment | added | John Dawkins | 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. | |
| Jan 2, 2022 at 15:42 | vote | accept | gigalord | ||
| Jan 2, 2022 at 14:06 | comment | added | gigalord | Thank you! I am not sure I follow the last step though. What do you mean by "usual simple function approximation"? | |
| Jan 1, 2022 at 18:18 | comment | added | Alex M. | Oh, then it's probably $\int _\Omega H \, \mathrm d [f(x, \cdot)]$. | |
| Jan 1, 2022 at 18:15 | comment | added | John Dawkins | 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 $\int_\Omega H(\omega) f(x,d\omega)$. | |
| Jan 1, 2022 at 18:04 | comment | added | Alex M. | 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. | |
| Jan 1, 2022 at 17:56 | history | answered | John Dawkins | CC BY-SA 4.0 |