I just started learning measure and rigorous integration theory on my own along side my calculus class and I've noticed that with the substitution rule, you have something that looks like this $$ \int^{b}_{a} f(g(x))g'(x) \, dx=\int^{u(b)}_{u(a)} f(u) \, du $$ where $u(x)=g(x)$. I don't know how to formally connect this with measure theory yet but when I was listening to the lecture in class, my instincts/intuition were screaming at me that this has to connect directly to measure theory. I know the statement of the Radon-Nikodym theorem, though I haven't been able to quite grasp the significance but that would be my first guess but a google search only yielded some feint allusions to this connection that were rather unsatisfactory. I just feel like the change of the bounds from $[a,b]$ to $[u(a),u(b)]$ has to be some sort of "change of measure" or something to that effect. I apologize if this is a very elementary or stupid question but I can't get it off my mind and I'm not making much progress trying to formalize it myself. Any ideas or just some hints that will point me in the right direction would be awesome. Thanks in advance.


Your instinct is on track. The "substitution rule" is related to a sort of change of measure, that of "image measure". Let's suppose that $g:I\to J$ is a strictly increasing $C^1$ of $I$ onto $J$, where $I$ and $J$ are closed intervals contained in $\Bbb R$. Let $f:J\to\Bbb R$ be integrable, and define a measure $\mu$ by the formula $\mu(B)=\int_B f(u)\,du$ for each Borel set $B\subset J$. Also define another measure $\nu$ for Borel subsets $A$ of $I$ by the formula $$ \nu(A):=\int_A f(g(x))g'(x)\,dx. $$ The substitution formula you cite amounts to the statement that $\nu([a,b])=\mu([g(a),g(b)]$ for each closed interval $[a,b]\subset I$. Noting that $g^{-1}([g(a),g(b)])=[a,b]$, this is the same as saying that $$ \mu(B)=\nu(g^{-1}(B)) $$ for each closed subinterval of $J$. From this we can deduce that $\mu(B)=\nu(g^{-1}(B))$ for all Borel subsets of $J$; this situation is often described by saying that $\mu$ is the image measure of $\nu$ under the transformation $g$. The image measure construction only requires a Borel measurable $g:I\to J$, but the explicit formula relating $\mu$ and $\nu$ as in the first display above requires some differentiability of $g$.

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    $\begingroup$ +1! Just beat me to it but this is an excellent, excellent answer..not sure if it helps but for $u:X \to B$, as an intuitive metaphor, I've always thought of a pushforward/image measure $\mu_{u}:\mathcal{F}_{2} \to [0,\infty]$ where $\mu_{u}(B)=\mu(u^{-1}(B))$ as a "bridge" between two measurable spaces $(\Omega_{1},\mathcal{F}_{1})$ and $(\Omega_{2},\mathcal{F}_{2})$. $\endgroup$ – Wavelet Jul 19 '16 at 16:28
  • $\begingroup$ I really appreciate your answer @JohnDawkins, it is more helpful than you probably realize. I'm also curious as to whether this has any connection to the Lebesgue-Steltjes (sp?) integral. I've been trying work through that the past few days but I couldn't understand the $\int_{A} f dg(x)$ for $g \in BV(A)$ notation. Does the $dg(x)$ really mean integration with respect to the image measure of $g$? $\endgroup$ – AnalysisStudent Jul 19 '16 at 20:39
  • $\begingroup$ @Wavelet, I see what you mean about thinking of it as a bridge between the two spaces. Or maybe a "rosetta stone" that allows one to "translate" a measure on the image of a measurable function so that it agrees with a measure on the pre-image of the function? Good thoughts though, I really can't thank you guys enough for the insights! $\endgroup$ – AnalysisStudent Jul 19 '16 at 20:44

Yes, your guess is correct. If $g: [a,b] \to g([a, b])$ is a diffeomorphism and $\lambda$ is the Lebesgue measure on $[a,b]$, then you may consider the push-forward of $\lambda$ through $g$, denoted $g_* \lambda$. The push-forward is defined by

$$(g_* \lambda) (A) = \lambda (g^{-1} (A))$$

for every measurable $A$.

This can be rewritten using integrals instead of measures as

$$\int _A 1 \ \Bbb d (g_* \lambda) = \int _{g^{-1} (A)} 1 \ \Bbb d \lambda ,$$

or equivalently

$$\int 1_A \ \Bbb d (g_* \lambda) = \int 1_{g^{-1} (A)} \ \Bbb d \lambda .$$

If $B = g(A)$, the preceding formula becomes

$$\int 1_{g^{-1} (B)} \ \Bbb d (g_* \lambda) = \int 1_B \ \Bbb d \lambda .$$

Notice now that

$$1_{g^{-1} (B)} (x) = \begin{cases} 1, & x \in g^{-1}(B) \\ 0, & x \notin g^{-1}(B) \end{cases} = \begin{cases} 1, & g(x) \in B \\ 0, & g(x) \notin B \end{cases} = 1_B (g(x)) = (1_B \circ g) (x), $$

so we may rewrite the above equality as

$$\int (1_B \circ g) \ \Bbb d (g_* \lambda) = \int 1_B \ \Bbb d \lambda .$$

Remembering that integrable functions are limits of step functions, the above leads to

$$\int (f \circ g) \ \Bbb d (g_* \lambda) = \int f \ \Bbb d \lambda$$

for every integrable function $f$.

Finally, it can be shown that $(g_* \lambda) (A) = |g'| \ \lambda (A)$, so the above gets rewritten as

$$\int (f \circ g) \ |g' (A)| \ \Bbb d \lambda = \int f \ \Bbb d \lambda .$$

In particular, if $g$ is increasing (notice that $g$ must be strictly monotonic, because $g' \ne 0$ by virtue of $g$ being a diffeomorphism), and using that the Lebesgue integral is just the Riemann integral for compact intervals (i.e. $\int _{[a,b]}$ is just the usual $\int _a ^b$), one obtains

$$\int \limits _a ^b f (g (x)) g' (x) \ \Bbb d x = \int \limits _{g(a)} ^{g(b)} f (x) \ \Bbb d x .$$

  • $\begingroup$ The detail of your answer makes it obvious that you have a ton of insight to offer on the subject and I really can't thank you enough for helping me connect the dots. $\endgroup$ – AnalysisStudent Jul 19 '16 at 20:54
  • $\begingroup$ @AnalysisStudent: Notice that the crucial formula $(g_* \lambda) (A) = |g'| \ \lambda (A)$ is not easy to prove, therefore I have not attempted to include its proof here. Also, remember that all of the above can easily be restated in $\Bbb R ^n$ (and even on smooth manifolds), with minimal modifications (the most important one is the replacement of $g'$ with $\det \ \Bbb d g$). $\endgroup$ – Alex M. Jul 19 '16 at 20:59
  • $\begingroup$ I'll have to take a closer look at it when I get home tonight but intuitively, I can see how that might be true. It seems that for measure of the image of $g(A)$ relative to the measure of pre-image $g^{-1}(A)$ depends on the "steepness" of the function $g$. If $g$ is really steep then a set with small measure $g^{-1}(A)$ can map to a set with a potentially very large measure (regardless of the sign of $g'$, hence the $\lvert g'\rvert)$. Just to sketch an example, let $g(x)=8x$ and $B=[g(0),g(1)]$ then $\lambda(g^{-1}(B))=1$ but $\lambda(B)=8$. Am I on the right track? $\endgroup$ – AnalysisStudent Jul 19 '16 at 21:32
  • $\begingroup$ Actually, that might only be true if $g$ is monotone. I'll work on it some more tonight and edit my question with what I come up with. $\endgroup$ – AnalysisStudent Jul 19 '16 at 21:35
  • $\begingroup$ @AnalysisStudent: 1) $g$ is always monotone if it is a diffeomorphism, because $g' \ne 0$ so it has to keep a constant sign. 2) Your intuitive understanding is correct in $\Bbb R$, but betrays you in $\Bbb R^n$, where $g'$ gets replaced by $\det g'$. Think of the linear maps of the form $g(x) = \begin{pmatrix} a & 0 \\ 0 & \frac 1 a \end{pmatrix} x$ with $a>0$ in $\Bbb R^2$: if $a$ is "far" from $1$ the distortion is big, yet the factor $\det g'$ doesn't catch that, being always $1$, so no, it's not exactly about distortion anymore. $\endgroup$ – Alex M. Jul 20 '16 at 6:32

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