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:
- Was my proof correct?
- Was my proof necessary? Am I thinking about this poorly?
Thanks!
Best Answer
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.