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We have the function $f:\mathbb R\to\mathbb R$:
$$f(x) =\begin{cases} y^{-2} & \text{if } 0\lt x\lt y \le 1 \\ -x^{-2} & \text{if } 0\lt y\lt x \le 1 \\ 0 & \text{otherwise } \end{cases} $$

Calculate the double integrals $$\int_0^1\bigg(\int_0^1f(x,y)\lambda(dx)\bigg)\lambda(dy)$$ and $$\int_0^1\bigg(\int_0^1f(x,y)\lambda(dy)\bigg)\lambda(dx)$$

Is $f\in\mathcal L^1(\lambda_2)$?

I'm not really sure how to approach this. I know how to calculate "normal" double integrals, but the fact that $f(x)$ changes according to whether $x$ or $y$ is biggest confuses me, so I don't really know how I should do it.

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Is $f$ a function of one or two variables? – copper.hat Dec 14 '12 at 9:16
up vote 1 down vote accepted

I suppose that you meant $f:\mathbb{R}^2\rightarrow\mathbb{R}\ , f=f(x,y)$ (I am positive that you mean that since the question is if $f\in\mathcal{L}^1(\lambda_2)$, that is if is a (Lebesgue) integrable function in $\mathbb{R}^2$). We have that:


Then: $\int_0^1\frac{2}{y}\lambda(dy)=2\int_0^1y^{-1}\lambda(dy)=2\int_0^1(\log y)'\lambda(dy)=2(\log1-\lim_{t\rightarrow0^+}\log t)=+\infty$

In the same way:



$\int_0^1(-x^3+\frac{1}{x}-1)\lambda(dx)=-\int_0^1x^3\lambda(dx)+\int_0^1\frac{1}{x}\lambda(dx)-\int_0^1\lambda(dx)=-\int_0^1\left(\frac{x^4}{4}\right)'\lambda(dx)+\int_0^1(\log x)'\lambda(dx)-1=-(\frac{1}{4}-0)+(\log1-\lim_{t\rightarrow0^+}\log t)-1=-\frac{1}{4}-1+\infty=+\infty$

Since the integrals $\int_0^1\left(\int_0^1f(x,y)\lambda(dx)\right)\lambda(dy)=+\infty$ and $\int_0^1\left(\int_0^1f(x,y)\lambda(dy)\right)\lambda(dx)=+\infty$, we have that f is not integrable, that is $f\notin \mathcal L^1(\lambda_2)$

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Split the inner integral into two pieces: $$\int_0^1f(x,y)dx=\int_0^yf(x,y)dx+\int_y^1f(x,y)dx$$ and $$\int_0^1f(x,y)dy=\int_0^xf(x,y)dx+\int_x^1f(x,y)dx$$

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But what should f(x,y) be? Given that it changes? Should it be $\int_0^y -x^{-2}dx + \int_y^1 y^{-2}dx$ in the top one, and then the other way around in the other one? – Mai Dec 14 '12 at 10:23
No, by the way you wrote $f(x,y)$, you get $$\int_0^yy^{-2}dx+\int_y^1-x^{-2}dx$$ in the first one – Dennis Gulko Dec 14 '12 at 11:03

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