# Well-definedness of improper integral that converges.

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!

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.