Abram Boyd

2022-06-24

Given the step function
$g\left(x,y\right)=\left\{\begin{array}{ll}\frac{1}{{x}^{2}}& 0
I want to show that
${\int }_{\left(0,1\right)}{\int }_{\left(0,1\right)}g\left(x,y\right)\phantom{\rule{thinmathspace}{0ex}}d\lambda \left(x\right)\phantom{\rule{thinmathspace}{0ex}}d\lambda \left(y\right)\ne {\int }_{\left(0,1\right)}{\int }_{\left(0,1\right)}g\left(x,y\right)\phantom{\rule{thinmathspace}{0ex}}d\lambda \left(y\right)\phantom{\rule{thinmathspace}{0ex}}d\lambda \left(x\right)$
However, when trying to use the relation with Riemann integrability I end up with divergent integrals such as
${\int }_{0}^{x}{\int }_{0}^{1}\frac{1}{{x}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy-{\int }_{0}^{1}{\int }_{0}^{y}\frac{1}{{y}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy$
I also considered using limits such as following one, but again I end up with divergence.
$\underset{k\to \mathrm{\infty }}{lim}{\int }_{\left(\frac{1}{k},1\right)}\underset{k\to \mathrm{\infty }}{lim}{\int }_{\left(\frac{1}{k},1\right)}g\left(x,y\right)\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy$
I would really appreciate any help!

Josie Stephenson

$\begin{array}{rl}& {\int }_{\left(0,1\right)}\left({\int }_{\left(0,1\right)}g\left(x,y\right)\phantom{\rule{thinmathspace}{0ex}}d\lambda \left(x\right)\right)\phantom{\rule{thinmathspace}{0ex}}d\lambda \left(y\right)\\ =& {\int }_{\left(0,1\right)}\left({\int }_{\left(0,y\right)}-\frac{1}{{y}^{2}}\phantom{\rule{thinmathspace}{0ex}}d\lambda \left(x\right)+{\int }_{\left(y,1\right)}\frac{1}{{x}^{2}}\phantom{\rule{thinmathspace}{0ex}}d\lambda \left(x\right)\right)\phantom{\rule{thinmathspace}{0ex}}d\lambda \left(y\right)\\ =& {\int }_{\left(0,1\right)}\left(-\frac{1}{y}+\frac{1-y}{y}\right)\phantom{\rule{thinmathspace}{0ex}}d\lambda \left(y\right)={\int }_{\left(0,1\right)}-1\phantom{\rule{thinmathspace}{0ex}}d\lambda \left(y\right)=-1.\end{array}$