# Measure theory analysis

#### HeinzBor

Problem Statement
Seeing if a funkcion is integrable
Relevant Equations
Fubinis theorem and Tonelli's theorem
Hi I am sitting with a homework problem which is to show if I can actually integrate a function. with 2D measure of lebesgue. the function is given by $\frac{x-y}{(x+y)^2} d \lambda^2 (x,y)$.

I know that a function $f$ is integrable if $f \in L^{1}(\mu) \iff \int |f|^{1} d \mu < \infty$.

Since $(f \geq 0)$ I can apply Tonelli's Theorem, which states that

$\int_{X \times E} f d_{\mu \times v} = \int_{X}(\int_{Y}fdv)d \mu = \int_{Y}(\int_{X}fd \mu)d v$

So my first idea was to compute both RHS and LHS and show that they do not equal if they are not measurable. But I saw that this was a complicated integral, so I was wondering if there is some other way to do it?

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#### fresh_42

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2018 Award
It is not so complicated if you split it: $\dfrac{x-y}{(x+y)^2} = \dfrac{x}{(x+y)^2} - \dfrac{y}{(x+y)^2}$.

#### HeinzBor

It is not so complicated if you split it: $\dfrac{x-y}{(x+y)^2} = \dfrac{x}{(x+y)^2} - \dfrac{y}{(x+y)^2}$.
Thanks and then integrate both sides of tonelli's theorem ? and if they are not equal then it is not integrable because of fubini?

#### Mark44

Mentor
From your deleted post:
$\int_{[0,1] Y} \frac{x}{(x+y)^2}dv(y) - \int_{[0,1] X} \frac{y}{(x+y)^2}d \mu (x)$
Please use either $or  delimiters on your LaTeX stuff. #### HeinzBor Again, single  delimiters don't do anything at this site. Please use either$ or  delimiters on your LaTeX stuff.
Sorry it should be fixed now

#### Mark44

Mentor
Your result looks fine to me. The integrand function is defined everywhere except at (0, 0) on the square $[0, 1] \times [0, 1]$, so the result via Lebesque integration should be the same as with ordinary (Riemann) integration. By wolframalpha, I get a value of 0 for the integral for either order of integration.

#### HeinzBor

Your result looks fine to me. The integrand function is defined everywhere except at (0, 0) on the square $[0, 1] \times [0, 1]$, so the result via Lebesque integration should be the same as with ordinary (Riemann) integration. By wolframalpha, I get a value of 0 for the integral for either order of integration.
okay thanks, so I can conclude that the integral exists?

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