The assumptions of the substitution theorem in double integral

Question

Let $$ \mathrm{T}: \left\{ \begin{array}{l} u=u(x,y) \\ v=v(x,y) \end{array} \right. $$ be a change of variables substitution, where $u$ and $v$ have continues partial derivative in an open set $D$. In addition we assume that the Jacobian $$ \frac{\partial(x,y)}{\partial(u,v)} $$ of $T$ does not vanish in $D$ and that $f(x,y)$ is integlable in the image $T(D)$. Under these condition I know that the formula of substitution in double integral is $$ \iint_{\mathrm{T(D)}}f(x,y)\,dxdy =\iint_{\mathrm{D}}f(x(u,v),y(u,v))\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,dudv $$ Am I missing any condition on the boundary of the set $D$? What should we assume further on the set $D$? Should it be connected? bounded? Thanks!

Answer 1

The usual way that change of variables is stated is

Theorem (Rogawski Fundamentals of Calculus $\S$16.6) Let $G: \mathcal{D}_0 \to \mathcal{D}$ be a $C^1$ mapping that is one-to-one on the interior of $\mathcal{D}_0$. If $f(x,y)$ is continuous, then $$\iint_\mathcal{D} f(x,y)~\mathrm{d}x~\mathrm{d}y = \iint_{\mathcal{D}_0} f(x(u,v), y(u,v)) \left\vert \frac{\partial (x,y)}{\partial (u,v) } \right \vert ~\mathrm{d} u ~\mathrm{d}v.$$

Stating that the partials are continuous gives that your map $T$ is $C^1$ (continuously differentiable). Stating that the Jacobian is non-vanishing gives that $T$ is locally invertible and therefore locally one-to-one. However, this does not guarantee that $T$ is one-to-one on the interior of $\mathcal{D}$.

For example, let $\mathcal{D}_1 = [0,1] \times [0,1]$, $\mathcal{D_2} = [2,3] \times [0,1]$, and $\mathcal{D}= \mathcal{D}_1 \cup \mathcal{D}_2$. Define $T=T(u,v)$ by $$T(u,v) = \begin{cases} (u,v) & (u,v) \in \mathcal{D}_1 \\ (u-2,v) & (u,v) \in \mathcal{D}_2\end{cases}.$$ Then the Jacobian of $T$ is $I_{2 \times 2}$ for all $(u,v) \in \mathcal{D}$ and $T$ is locally invertible, but $T$ is two-to-one on $\mathcal{D}$. As such, you would need to add a topological restriction to $\mathcal{D}$ in order to guarantee that $T$ will be one-to-one.

As for the boundary of $\mathcal{D}$, there is no restriction that $T$ be one-to-one there. For example, the polar coordinates transformation $$x = r \cos \theta \\ y = r \sin \theta$$ on $[0, R] \times [ 0, 2 \pi]$ is infinity-to-one on $\{0\} \times[0, 2 \pi]$ and two-to-one on $[0,R] \times \left( \{0\} \cup \{2 \pi\} \right)$, but this introduces no issues as these are sets of measure zero whose images are also of measure zero.

The quoted theorem being from a multi-variable calculus text, there are many standing assumptions---namely, that a domain is a closed set with piecewise smooth boundary. If we wish to generalize this to other domains $\mathcal{D}$, the key piece is that $\partial \mathcal{D}$ is of measure zero (generally difficult to violate for reasonable domains). For more information, see Is there a change of variables formula for a measure theoretic integral that does not use the Lebesgue measure.