Abram Boyd

2022-06-24

Given the step function

$g(x,y)=\{\begin{array}{ll}\frac{1}{{x}^{2}}& 0<y<x<1\\ -\frac{1}{{y}^{2}}& 0<x<y<1\\ 0& \text{otherwise}\end{array}$

I want to show that

${\int}_{(0,1)}{\int}_{(0,1)}g(x,y)\phantom{\rule{thinmathspace}{0ex}}d\lambda (x)\phantom{\rule{thinmathspace}{0ex}}d\lambda (y)\ne {\int}_{(0,1)}{\int}_{(0,1)}g(x,y)\phantom{\rule{thinmathspace}{0ex}}d\lambda (y)\phantom{\rule{thinmathspace}{0ex}}d\lambda (x)$

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.

$\underset{k\to \mathrm{\infty}}{lim}{\int}_{(\frac{1}{k},1)}\underset{k\to \mathrm{\infty}}{lim}{\int}_{(\frac{1}{k},1)}g(x,y)\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy$

I would really appreciate any help!

$g(x,y)=\{\begin{array}{ll}\frac{1}{{x}^{2}}& 0<y<x<1\\ -\frac{1}{{y}^{2}}& 0<x<y<1\\ 0& \text{otherwise}\end{array}$

I want to show that

${\int}_{(0,1)}{\int}_{(0,1)}g(x,y)\phantom{\rule{thinmathspace}{0ex}}d\lambda (x)\phantom{\rule{thinmathspace}{0ex}}d\lambda (y)\ne {\int}_{(0,1)}{\int}_{(0,1)}g(x,y)\phantom{\rule{thinmathspace}{0ex}}d\lambda (y)\phantom{\rule{thinmathspace}{0ex}}d\lambda (x)$

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.

$\underset{k\to \mathrm{\infty}}{lim}{\int}_{(\frac{1}{k},1)}\underset{k\to \mathrm{\infty}}{lim}{\int}_{(\frac{1}{k},1)}g(x,y)\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy$

I would really appreciate any help!

Josie Stephenson

Beginner2022-06-25Added 20 answers

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

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