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I came across the following statement in Lehmann & Romano's Testing Statistical Hypotheses:

Lemma 2.3.2 Let $T:(\mathcal{X},\mathcal{A})\to(\mathcal{T},\mathcal{B})$ be a measurable transformation, $\mu$ a $\sigma$-finite measure over $(\mathcal{X},\mathcal{A})$, and $g:(\mathcal{T},\mathcal{B})\to(\mathbb{R},\mathfrak{B}(\mathbb{R}))$ a real-valued measurable function. If $\mu^{*}$ is the measure defined over $(\mathcal{T},\mathcal{B})$ by $$ \mu^{\ast}(B)=\mu[T^{-1}(B)],\quad\forall B\in\mathcal{B}, $$ then $$ \int_{T^{-1}(B)}g\circ T\,\mathrm{d}\mu=\int_{B}g\,\mathrm{d}\mu^{\ast},\quad\forall B\in\mathcal{B}, $$ in the sense that if either integral exists, so does the other and the two are equal.

Can we make the same statement without requiring $\mu$ to be $\sigma$-finite? And why not?

Instead of providing a complete proof, the authors suggest to go through the standard generalisation procedures from simple to non-negative to real-values measurable functions. Other texts I have access to either state $\sigma$-finiteness as a standing assumption or define Lebesgue integration for the Borel-Lebesgue measure only (which is $\sigma$-finite). I couldn't see where I would rely on $\sigma$-finiteness in my transition from simple to general measurable functions. To be honest, neither could I find where $\sigma$-finiteness is used in the Lebesgue integral definition, and other people seem to suffer from the same lack of clarity, judging from these unanswered questions:

I know that $\sigma$-finiteness is necessary for the existence of the Radon-Nikodym derivative, and counter-example are known and discussed here:

I am looking for a similar example for the above statement or an explanation of the exact role of $\sigma$-finiteness in the change-of-variables lemma.

Hints and suggestions are appreciated, but nothing is better than a complete answer!

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    $\sigma$-finiteness is redundant for this abstract change of variables theorem. – peek-a-boo Nov 02 '22 at 23:28
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    The change of variable in your Lemma does not require $\sigma$-finiteness. The push forward $\nu(f^{-1}(\cdot))$ works only because presages behave well under complementation, and unions (and intersections). – Mittens Nov 02 '22 at 23:28
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    ... meant to say preimages. – Mittens Nov 03 '22 at 00:15

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