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Let's say I have a function $f:\mathbb R^2\longrightarrow\mathbb R$ given by $$ f(x,y) = \begin{cases} g(x,y)\quad 0<x<y<\infty\\0 \quad\text{else}\end{cases}$$ and I want to calculate the Lebesgue integral over this function (assuming all necessary conditions regarding $g$ are met for integration). $$ $$ Assuming that $g$ is non-negative, I can use Tonellis Theorem to get $$ \int_{\mathbb R^2} f\ d(x,y) = \int_\mathbb R\int_\mathbb R ?$$ but then I am stuck because I don't know how to split up the indicator to something like $1_A(x)$ and $1_B(y)$ where $A$ and $B$ are sets associated to the integration bounds. If I set $A=]0,y[$ and $B=]x,\infty[$ I have the problem the indicator depends on the one variable and the corresponding sets and the other. How do I correctly integrate over such an indicator function?

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For fixed $y$, let $g^y(x)=g(x,y)$. Then $$\int_{\mathbb{R}^2}fd(x,y)=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}g^y\chi_{[0,y)}dm(x)\right)\chi_{[0,\infty)}dm(y)$$

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For fixed $x$, let $g_x(y)=g(x,y)$. Then $$\int_{\mathbb{R}^2}fd(x,y)=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}g_x\chi_{[x,\infty)}dm(y)\right)\chi_{[0,\infty)}dm(x)$$

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  • $\begingroup$ Thank you for your reply. Before I accept it: can you please say what would be the correct indicators when integrating with respect to $y$ first and $x$ second? $\endgroup$ – Syd Amerikaner Jul 17 '17 at 19:31
  • $\begingroup$ @SydAmerikaner Answer edited $\endgroup$ – tattwamasi amrutam Jul 17 '17 at 19:37
  • $\begingroup$ Thank you! I believe I understand it now. $\endgroup$ – Syd Amerikaner Jul 17 '17 at 19:40

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