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Measure theory analysis

  • Thread starter HeinzBor
  • Start date
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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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}##.
 
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?
 
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From your deleted post:
$\int_{[0,1] Y} \frac{x}{(x+y)^2}dv(y) - \int_{[0,1] X} \frac{y}{(x+y)^2}d \mu (x)$
For MathJax on this site, the delimiters are a pair of dollar signs ($) for standalone formulas) or a pair of hash signs (#) for inline formulas).
Using the latter, the integral above is ##\int_{[0,1] Y} \frac{x}{(x+y)^2}dv(y) - \int_{[0,1] X} \frac{y}{(x+y)^2}d \mu (x)## - that's with two # characters at the start and two more at the end.
 
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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}##.
Another possibility is ##\dfrac{x-y}{(x+y)^2} = \dfrac{x + y}{(x+y)^2} - \dfrac{2y}{(x+y)^2}##.
Also, from post #1, a minor point:
the function is given by ##\frac{x-y}{(x+y)^2} d \lambda^2 (x,y)##.
The integrand is just this part: ##\frac{x - y}{(x + y)^2}##.
 
Okay thank you

Okay so I gave it a try


\begin{align*}

\int_{[0,1] \times [0,1]} \frac{x-y}{(x+y)^2} d \lambda^2 (x,y)

\end{align*}

Since $$f \geq 0$$ from Tonelli we have that

\begin{align*}

\int_{[0,1]} (\int_{[0,1]} \frac{x-y}{(x+y)^2} d \lambda(x)) \ d \lambda (y)\\

= \int_{[0,1]} (\int_{[0,1]} \frac{x+y}{(x+y)^2} - \frac{2y}{(x+y)^2} d \lambda(x)) \ d \lambda (y)

\end{align*}

and since $$\frac{2y}{(x+y)^2}$$ is maximum 2 we can just see it as a finite constant and then consider

\begin{align*}

&\leq \int_{[0,1]} (\int_{[0,1]} \frac{x-y}{(x+y)^2} d \lambda(x)) d \lambda (y)\\

&= \int_{[0,1]} \frac{1}{2}ln((y+1)^2) - \frac{1}{2}ln(y^2) d \lambda (y)\\

&= 2ln(2) < + \infty

\end{align*}

so f is integrable. is this correct? I hope you can help me if theres some mistakes I am still trying to learn more about the integral and measure
 
Last edited:
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
 
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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.
 
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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