0
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.