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I am reading Rudin's proof of the change of variable theorem (theorem 10.9 in baby rudin). $$\int_{R^k}{f(y)dy}=\int_{R^k}{f(T(\mathbf{x}))\left\lvert J_T(\mathbf{x})\right\rvert d\mathbf{x}}$$ I am having trouble understanding how he proves that the change of variable theorem holds true in the case that the transformation between coordinates is a primitive C' mapping. It seems like the idea is that in the primitive mapping case, this is pretty much equivalent to the already-proven single variable substitution rule. He states that $$\int_{R^1}{f(y)dy}=\int_{R^1}{f(T(x))T'(x) dx}$$ is true when $T(x)$ is increasing, as a case of the single variable substitution rule, but that the absolute value of $T'(x)$ is needed in order to make the relation hold if $T(x)$ is decreasing. This seems like a contradiction to me, because it is well known that the single variable substitution rule does not depend on T(x) being increasing, and this inclusion of absolute value changes that theorem from calculus 1. I think that this has something to do with the fact that the integral is over $R^1$, and the way he has defined the integral in this chapter as an integral over any k-cell that contains the compact support of f. If anyone can further help me that is familiar with this proof, I would greatly appreciate it!

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    $\begingroup$ The issue is that the rule we are taught in calculus has a 'signed' area, we have $\int_a^b = - \int_b^a$, but when we write $\int_A$ there is no notion of orientation/direction. That is, with the latter, if we use the change of variable $T(x) = -x$, we have $\int_{\mathbb{R}} f = \int_{\mathbb{R}} (f \circ T)|T'| $, but $\int_{-\infty}^\infty f = \int_{\infty}^{-\infty} (f \circ T) T'$. $\endgroup$
    – copper.hat
    Aug 6 '14 at 23:44
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Take $\int_{\bf R^1} f(y)\operatorname{d}y$ and substitute using any strictly decreasing surjective function; for example: $y=-x$

$$\begin{align} \int_{\bf R^1} f(y)\operatorname{d}y & = \int_{-\infty}^{+\infty} f(y)\operatorname{d}y \\[1ex] & = \int_{+\infty}^{-\infty} f(-x)\cdot(-1)\operatorname{d}x & \text{note the change of limits} \\[1ex] & = \int_{-\infty}^{+\infty} f(-x)\operatorname{d}x & \text{since: } -\int_a^b F(x) \operatorname{d}x = \int_b^a F(x)\operatorname{d}x \\[2ex] & = \int_{\bf R^1} f(-x) \,\left|\frac{\operatorname{d}(-x)}{\operatorname{d}x}\right|\,\operatorname{d} x \end{align}$$ This will hold for any substitution that is surjective and the integration is over the domain of the integration.

If such a $y=T(x)$ is strictly decreasing, then by definition: the derivative will be negative and the the domain $x$ will run in the opposite direction.  So using the absolute of the derivative ensures that the direction of the integration remains the same.

$$\int_{\bf R^1} f(y)\operatorname{d}y = \int_{\bf R^1} f(T(x)) \,|T'(x)|\,\operatorname{d}x$$

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  • $\begingroup$ thank you very much. I had a feeling this was the problem, and your comment about the surjectivity of the function T(x) really clarified this for me. $\endgroup$
    – nickodel
    Aug 7 '14 at 5:20

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