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In measure theory we learn that $$ \int_\Omega g \circ f d\mu = \int_{f(\Omega)}g d(\mu \circ f^{-1}) $$ where $(\Omega, \mathscr F, \mu)$ is a measure space and $f$ and $g$ are measurable.

Now in calculus we have that $$ \int_{\phi(a)}^{\phi(b)} h(x)dx = \int_a^b h(\phi(t))\phi'(t)dt $$ for a suitable substitution $x = \phi(t)$. From this substitution we have $dx = \phi'(t)dt$ with $\phi'(t)$ being the Jacobian, but I want to understand this in terms of the measure theoretic formulation.

Clearly I'll have $g = h$ and $f = \phi$, so this ought to be equivalent to $$ \int_{[a,b]} h \circ \phi d\mu \stackrel ?= \int_{\phi([a,b])} h \ d(\mu \circ \phi^{-1}). $$

Let $\lambda$ be the Lebesgue measure. It seems fair to assume that $\mu \ll \lambda$ so the Radon-Nikodym derivative $\frac{d\mu}{d\lambda}$ exists. This means that $$ \int_{[a,b]}h \circ \phi d\mu = \int_{[a,b]} h \circ \phi \frac{d\mu}{d\lambda}d\lambda $$ so we have $\phi' = d\mu / d\lambda$?

Now how do we reconcile this with the $\mu \circ \phi^{-1}?$

alfalfa
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    IMO, one of the biggest awkward points about doing calculus with measures is that they are notated like differentials, but really don't behave like them; e.g. they act in the opposite way as a differential does in regards to a change of variable. –  Jul 07 '17 at 02:23
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    @Hurkyl i'm glad to hear other people feel that way. I feel like reconciling the notation with the ideas is harder than the actual ideas, at least for basic stuff like this – alfalfa Jul 13 '17 at 20:27

2 Answers2

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Symbolically we have $\phi'(t) \, dt = d\phi$ which gives $$ \int_a^b (h \circ \phi)(t) \, \phi'(t) dt = \int_{[a,b]} (h \circ \phi)(t) \, d\phi(t) = \int_{\phi([a,b])} h(x) \, d\phi(\phi^{-1}(x)) = \int_{\phi(a)}^{\phi(b)} h(x) \, dx $$

The measure $\mu$ in your post is defined by $\mu(E) = \int_E \phi'(t) \, dt = \int_E \phi'(t) \, d\lambda(t)$ so $$\frac{d\mu}{d\lambda}(t) = \phi'(t).$$

md2perpe
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    thanks, this is exactly what i was looking for! – alfalfa Jul 06 '17 at 16:35
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    I guess you need to say that $\phi $ is strictly increasing; otherwise $\phi([a,b])$ may not be equal to $[\phi(a),\phi(b)]$. I also think the claim that $d\phi (\phi^{-1}(x))=dx$ needs to be justified measure-theoretically. – Matematleta Jul 06 '17 at 23:43
  • @ChilangoIncomprendido. Sure, but I think that it was actually the second part that alfalfa wanted.

    Symbolically, $d\phi(\phi^{-1}(x))$ is better justified if written $\phi(\phi^{-1}(dx))$. Of course it still needs to be justified measure-theoretically.

     – md2perpe Jul 07 '17 at 23:08
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Assume that $\phi $ is strictly increasing and define a measure $\mu$ by $\mu(E)=\int_E\phi'dx.$

Now take $h=\chi_{[\phi(a'),\phi (b')]}$ where $[\phi(a'),\phi (b')]\subseteq [\phi(a),\phi(b)].$ Then,

$\int^{\phi(b)}_{\phi(a)}\chi_{[\phi(a'),\phi (b')]}d\mu\phi^{-1}=\mu([a',b'])=\int_{a'}^{b'}\phi'dx=\phi(b')-\phi(a')=\int^{b}_{a}\chi_{[\phi(a'),\phi (b')]}\circ \phi dx$

so the formula is correct if $h$ is a characteristic function of an interval and therefore for $h\in L^1([a,b])$ as well, by the density of the step functions in $L^1([a,b]).$

Matematleta
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