What Is This Measure Theory Rule Called? Is it Just the Substitution Rule in Disguise? Let $(\Omega,\mathcal{F},\mu)$ be a measure space. For a $\mathcal{B}(\mathbb{R})$-measurable function $f \geq 0 : \Omega \rightarrow \mathbb{R}$ one can define a measure $\nu$ by:
$$
\nu(A) : = \int_A f d \mu = \int f \mathbb{1}_{A} d \mu
$$
$\nu$ is provably a measure according to this answer.
Now suppose we have some $\mathcal{B}(\mathbb{R})$-measurable function $g : \Omega \rightarrow \mathbb{R}$. And suppose $g$ is integrable w.r.t. $\nu$.
Question: Does this identity have a standard name in measure theory:
$$\int gf d \mu = \int g d \nu$$
Note: maybe the exact identity above isn't a named result in measure theory, but can be shown via the application of one or two standard measure theory results?

I think I can prove the identity using first principles of measure theory, but I'd rather use more standard results and my common sense is telling me the above is so simple it must have a standard name in measure theory that I'm just missing.
The proof of the identity I imagine is to use the approximation theorem for integrals, which asserts that a sequence of increasing simple functions approximates $\int g d \nu$. That is, we approximate with functions of the form:
$$\sum_{i=1}^n a_i \nu(A_i) = \sum_{i=1}^n a_i \int_{A_i} f d \mu$$
We can then invoke the approximation theorem again for the integrals inside the sum:
$$\sum_{i=1}^n a_i \sum_{j=1}^m b_{i,j} \mu(B_{i, j}) = \sum_{i=1}^n \sum_{j=1}^m a_ib_{i,j} \mu(B_{i, j})$$
Where $\bigcup_{j=1}^m B_{i, j} = A_i$. Since all the $B_{i,j}$ are disjoint, then the above is a simple function. Moreover, our invocations of the approximation theorem guarantee us that:

*

*$\forall \omega \in B_{i,j} \subseteq A_i : a_i \leq g(\omega)$

*$\forall \omega \in B_{i,j} : b_{i,j} \leq f(\omega)$
Therefore, we have:
$$\forall \omega \in B_{i,j} : a_ib_{i,j} \leq gf(\omega)$$
And so the term on the RHS is the integral of a simple function which is bounded by $gf$. And since the integral of $gf$ is the supremum over all such integrals of simple functions we have the bound:
$$\int gf d \mu \geq \int g d \nu$$
OK... so this is just an inequality rather than an identify. But I believe the proof can be fixed & completed by invoking the approximation theorem more fully: this will give us a sequence of simple functions converging to $f$ and another sequence converging to $g$. Then we can show that the resulting simple functions constructed above converges to $gf$ and so by the monotone convergence theorem the integrals will converge also.
But rather than complete this result in my own clunky way, I'd rather first see if there is some known result or results that can be applied here rather than go right back to first principles and simple functions.

Note: The induced measure $\nu$ is a measure on the original measurable space $(\Omega,\mathcal{F})$ - not a pushforward measure to $\mathcal{B}(\mathbb{R})$. So I can't see how something like change of variables / substitution can be applied. Unless I'm just not being creative enough...
Credit to this question for borrowed text.
 A: Thanks to the commenters for pointing me in the right direction. Since no formal answer has been posted I will do so on behalf of the commenters so that future readers may see the full argument. The crux of the argument is to use the Radon-Nikodym theorem, and crucially its uniqueness.
First we show that $\nu \ll \mu$. This follows from that fact that if $\mu(A) = 0$ then by the definition of the integral we have:
$$\nu(A) = \int_A f d \mu = 0 $$
Since $\nu \ll \mu$, then the Radon-Nikodym theorem is in force. This means that there exists a function $f'$ such that for all measurable sets $A$ the following identity holds:
$$\nu(A) = \int_A f' d \mu$$
Furthermore, the Radon-Nikodym theorem gives us that $f'$ is uniquely defined up to a $\mu$-null set, meaning that we have:
$$f = f' \;\;\; \mathrm{a.e.}$$
And so for any function $g$ we have:
$$gf = gf' \;\;\; \mathrm{a.e.}$$
Now, another property of any Radon-Nikodym derivative $f'$, as stated on Wikipedia, is that:
$$\int g d\nu = \int gf'd\mu$$
Putting it all together then:
$$\int g d\nu = \int gf'd\mu = \int gfd\mu$$
Which completes the proof.
