Assuming the usual required conditions (https://en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem#Radon%E2%80%93Nikodym_theorem), the Radon-Nikodym theorem states:

There exists a measurable function $f$ such that for every measurable $A\subseteq X$:
$v(A) = \int_{A} f d\mu$

However, to utilize this theorem, I find it useful to say:
$\int gdv = \int g fd\mu$ for every $g$ that is (?)-measurable.

A hand-wavy proof would be:
$v(A) = \int_{A} f d\mu (\forall \text{ measurable } A)$
$dv = fd\mu$
$gdv = gfd\mu$
$\int_A gdv = \int_A gfd\mu (\forall \text{ measurable } A)$

But I don't want a hand-wavy proof. I want a rigorous proof. To me, "$dv$" basically has no meaning. "$v$" is just a parameter to the integral, just like the integrand is a parameter to the integral. The "$d$" is only inserted to indicate that "$v$" is the second parameter to the integral, not to be confused with the first.

This is my attempt to make the theorem work for any measurable $g$:
First consider $g=\chi_B$ to be an indicator function on a measurable set $B$. Clearly,
$\int_A gdv = \int_{A\cap B} dv = \int_{A \cap B} fd\mu = \int_A gfd\mu (\forall \text{ measurable } A)$
Now, consider $g$ to be a simple function (finite sum of indicator functions). By linearity, our claim still holds.
Finally, consider $g$ to be the limit of a sequence of increasing simple functions. By MCT, our claim still holds.

My questions are:

  1. Was my proof correct?
  2. Was my proof necessary? Am I thinking about this poorly?


  • $\begingroup$ The measure $\nu$ with density $f$ with respect to $\mu$ is sometimes written $f\cdot\mu$. You might prefer it to $f d\mu$. $\endgroup$ Commented May 24, 2023 at 2:17

2 Answers 2


Your proof is not completely correct. You may refer to this Wikipedia page, and it actually requires $g$ to be integrable. In your notation, $g$ needs to be $\nu$-integrable.

The reason is that, in your last step, you use MCT. Note that, if a measurable function is not non-negative, you may not find a sequence of increasing simple functions to approximate the measurable function. Therefore, you need to write $g$ as $g^+ -g^-$ and your integral on the left hand side will need to be written as $\int_A g^+ d\nu -\int_A g^- d\nu$. Then you prove $\int_A g^+dv = \int_A g^+fd\mu$ and $\int_A g^-dv = \int_A g^-fd\mu$, respectively. You then may wish to subtract the second equation from the first equation to conclude the argument. But if you don't know $g$ is $\nu$-integrable, then you may have $\int_A g^+dv=\infty$ and $\int_A g^-dv=\infty$, and you cannot subtract the second equation from the first equation.


Your proof is correct. If you use the proof of Radon-Nikodym’s theorem from Wikipedia you indeed arrive at $$ \nu(A) = \int_A f d\mu $$ for measurable sets $A$, so your proof is also necessary if you want to say something about $\int gd\nu$ for $g$ measurable.

Edit: your proof suffices for $g$ non-negative, then for $g$ $\nu$-measurable. Without either of these hypotheses it fails.

  • $\begingroup$ Ah, I see this listed as one of bullet points in the "Properties" section of the wiki page now. Thanks!! $\endgroup$
    – Scott Hahn
    Commented May 24, 2023 at 15:27
  • 1
    $\begingroup$ The last step in OP's argument using MCT is incorrect. To be more specific, the last step in OP's argument may not apply to all measurable functions. It works for measurable functions which are $\nu$-integrable. $\endgroup$
    – Sam Wong
    Commented Jul 7, 2023 at 12:58

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