When I read Rudin's proof of change of variables,
I have a problem underlined in red below:
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!
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 $G(x)=y$, $$ \int_{\mathbb R^k} f(y)dy_1 \dots dy_k \\= \int_{\mathbb R^{k-1}} \int_{\mathbb R} F(\hat y,y_j) dy_j d\hat y \\= \int_{\mathbb R^{k-1}} \int_{\mathbb R} F(\hat x,y_j) dy_j d\hat x \qquad ({\hat x = \hat y})\\=\int_{\mathbb R^{k-1}} \int_{\mathbb R} F(\hat x,g(x)) |\partial_j g(x)| dx_j d\hat x \\= \int_{\mathbb R^k} f(G(x)) |J_G(x)| dx$$