What does $\int_{B_r} h$ mean? Problem 3-41 in "Calculus on Manifolds" by Michael Spivak. $\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}$.

Question

I am reading "Calculus on Manifolds" by Michael Spivak.

I solved (a) and (b).
Then, I proved $\int_C h=\int_{r_1}^{r_2} \int_{\theta_1}^{\theta_2} rg(r,\theta)d\theta dr$ holds.

My question is the following:

If $\theta_1:=0$ and $\theta_2:=2\pi$ and $r_1:=0$ and $r_2:=r$, then $C$ is an open set and $C\approx B_r$, but $C\neq B_r$.
$h$ is defined on $C\subset A$.
$\{(x,y):0\leq x\leq r\}\subset B_r$.
$\{(x,y):0\leq x\leq r\}\cap A=\varnothing$.
$h$ is not defined on $B_r$.
But the author says "show that $\int_{B_r} h=\int_{0}^{r} \int_{0}^{2\pi} rg(r,\theta)d\theta dr$".

What does $\int_{B_r} h$ mean?

Answer 1

If $\theta_1=0$, $\theta_2=2\pi$, $r_1=0$ and $r_2=r$, then the complement $B_r\setminus C$ is a set of measure $0$. This means that the definition of $h$ can be extended to $B_r$ in any way you want, and the integral over $B_r$ will not depend on how the function was extended.

In the theory of integration, it is common to integrate not actual functions, but equivalence classes of functions, where two functions are defined to be equivalent if they agree outside a measure $0$ set. See for example the discussion here: What does mean : element of $L^p$ are equivalence class rather than function.