When I read Rudin's proof of change of variables,
I have a problem underlined in red below:

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I don't understand why (31) is true when $T$ is a primitive $C'$-mapping.
I know that a primitive mapping is a translation along one coordinate.
But the translation function $g$ involves not one variable, but a vector!
Maybe this makes the problem difficult.
Please comment. Thanks!

  • $\begingroup$ Absolutely, that mapping $g$ what you give is the primitive. What I want to know that, why the change of variables holds if $T$ is 1-1 primitive $C'$-mapping, although we must ultimately show that the change of variables holds for general 1-1 $C'$-mapping. $\endgroup$
    – Hs P
    Feb 7 '19 at 12:10

Let $j$ be fixed and suppose $G(x)=\sum_{i\neq j} x_ie_i + g(x) e_j$ is a 1-1 primitive $C^1$ map satisfying the assumptions. Then $$ \partial_i G(x)= \begin{cases} e_i + \partial_i g(x) e_j & i\neq j\\ \partial_j g(x)e_j &i=j\end{cases}$$ So the matrix $\nabla G$ is the identity matrix with the $j$th row swapped for $\nabla g^T$. It follows $|J_G(x) | = |\partial_j g(x)|$. Now write for the $k-1$ dimensional vector $\hat y:= (y_1,\dots,y_{j-1},y_{j+1},\dots,y_k)$, $F(\hat y,y_j) := f(y)$. Using the 1D result and $G(x)=y$, $$ \int_{\mathbb R^k} f(y)dy_1 \dots dy_k \\= \int_{\mathbb R^{k-1}} \int_{\mathbb R} F(\hat y,y_j) dy_j d\hat y \\= \int_{\mathbb R^{k-1}} \int_{\mathbb R} F(\hat x,y_j) dy_j d\hat x \qquad ({\hat x = \hat y})\\=\int_{\mathbb R^{k-1}} \int_{\mathbb R} F(\hat x,g(x)) |\partial_j g(x)| dx_j d\hat x \\= \int_{\mathbb R^k} f(G(x)) |J_G(x)| dx$$

  • $\begingroup$ @HsP When integrating in $x_j$, the other components $\hat x$ are treated as a constant. $\endgroup$ Feb 7 '19 at 12:41
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    $\begingroup$ Aha! I got it! Thank for your kindeness! $\endgroup$
    – Hs P
    Feb 7 '19 at 12:45
  • $\begingroup$ (@HsP there was an error with my jacobian calculation...my apologies) $\endgroup$ Feb 7 '19 at 13:40
  • $\begingroup$ It might be worth mentioning that because of the assumption $J_T(\textbf{x})\neq 0$, in each primitive mapping $G$ we have $\partial_j g$ is always positive or always negative, for all $x$ in $E$. Otherwise $\partial_j g(x_0) = 0$ for some $x_0$, and then $J_G(x_0) = 0$ and $J_T$ vanishes in the corresponding argument. $\endgroup$ May 29 at 23:28

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