Well-definedness of improper integral that converges.

Question

I am computing a following integral $$ I = \int_{\| \mathbf{z} \| \leq 1 } f(\mathbf{z}) dz_1 dz_2 dz_3. $$ $\| \cdot\|$ is $L^2$ norm, $f \geq 0$, and $f(\mathbf{z})$ is infinity at $\mathbf{z} = (0,0,0)$. Now I can show that $$ \lim_{\varepsilon \to 0} \ ( \int_{-1}^{- \varepsilon} + \int_{\varepsilon}^1 )\int_{-\sqrt{ 1 - z_3^2}}^{\sqrt{ 1 - z_3^2}} \int_{ - \sqrt{ 1 - z_2^2 - z_3^2 } }^{ \sqrt{ 1 - z_2^2 - z_3^2 } } f(\mathbf{z}) d z_1 d z_2 d z_3 $$ converges. But how do I know that this value is the same value as if I define the limit in some other way? for example $$ \lim_{\varepsilon \to 0} \ ( \int_{-1}^{- \varepsilon} + \int_{\varepsilon}^1 )\int_{-\sqrt{ 1 - z_1^2}}^{\sqrt{ 1 - z_1^2}} \int_{ - \sqrt{ 1 - z_2^2 - z_1^2 } }^{ \sqrt{ 1 - z_2^2 - z_1^2 } } f(\mathbf{z}) d z_3 d z_2 d z_1 $$ or $$ \lim_{\varepsilon \to 0} \int_{ \varepsilon \leq \| \mathbf{z} \| \leq 1 } f(\mathbf{z}) dz_1 dz_2 dz_3 $$ Any explanation is appreciated. Thank you!

Answer 1

The monotone convergence theorem of measure theory comes to the rescue: You have described three domains of integration indexed by $\epsilon,$ let's call them $U_\epsilon,V_\epsilon,W_\epsilon.$ Your integrals, prior to taking the limit, can be written as

$$\int_B \chi_{U_\epsilon}f(z)\,dz,\,\int_B \chi_{V_\epsilon}f(z)\,dz,\,\int_B \chi_{W_\epsilon}f(z)\,dz.$$

Here $B=\{|x|\le 1\}.$ As $\epsilon\to 0,$ $\chi_{U_\epsilon}$ increases to $\chi_{B\setminus E},$ where $E$ is a set of volume $0.$ Same kind of thing for the other two characteristic functions. The MCT, which applies to positive functions like your $f,$ says that these limits are the same.