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I've put this question on math.SE for a while without getting any answers. I thought it must be a rather trivial question for MO so that I didn't put it here. But I do want to get some help anyway (before being closed for whatever reason).

In the setting of Riemann integration, we have the following change of variables formula:

Let $[a,b]$ be a closed interval, and let $\phi:[a,b]\to[\phi(a),\phi(b)]$ be a continuous monotone increasing function. Let $f:[\phi(a),\phi(b)]\to{\Bbb R}$ be a Riemann integrable function on $[\phi(a),\phi(b)]$. The $f\circ\phi:[a,b]\to{\Bbb R}$ is Riemann-Stieltjes integrable with respect to $\phi$ on $[a,b]$ and $$ \int_{[a,b]}f\circ\phi\ d\phi=\int_{[\phi(a),\phi(b)]}f. \tag{1} $$

In the setting of Lebesgue integration, we have the following:

Let $(X,{\mathcal B},\mu)$ be a measure space, and let $\phi:X\to Y$ be a measurable morphism from $(X,{\mathcal B})$ to another measurable space $(Y,{\mathcal C})$. Define the pushforward $\phi_*\mu:{\mathcal C}\to[0,+\infty]$ of $\mu$ by $\phi$ by the formula $\phi_*(E):=\mu(\phi^{-1}(E))$. If $f:Y\to[0,+\infty]$ is measurable, then $$ \int_Y f\ d\phi_*\mu=\int_X(f\circ\phi)\ d\mu. \tag{2} $$

My question is: how can I interpret (1) in terms of (2)?


A naive attempt is letting $X=[a,b]$ and $\phi$ be the measurable morphism from $[a,b]$ to $[\phi(a),\phi(b)]$ in (1). But the result doesn't match (2) at all.

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    $\begingroup$ I think the issue is that the $\phi$ in (1) is the inverse of the $\phi$ in (2). $\endgroup$ – Anthony Quas Mar 25 '13 at 14:51
  • $\begingroup$ @Anthony Quas, Thanks! So the rest of things would be calculating the pushforward. $\endgroup$ – Jack Mar 26 '13 at 2:54
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If $\phi:X\to Y$ is a $C^1$ diffeomorphism between open subsets of euclidean space, then the change of variables formula reads $$\int_Y f(y)\, d\lambda(y)=\int_X (f\circ\phi)(x)|\det\phi'(x)|\,d\lambda(x).$$ This means that the Lebesgue measure is the push-forward, $\lambda=\phi_* \mu$, of the absolutely continuous measure $\mu=|\det\phi'|\,\lambda$.

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