Consider the function $f$ on $[0,1]\times [0,1]$ given by

$$f(x,y) = \frac{x^2-y^2}{(x^2+y^2)^2}, \,(x,y)\neq (0,0)$$

and $f(0,0) = 0.$

Let $M$ denote the $\sigma$-algebra of Lebesgue measuable sets and $m$ the Lebesgue measure.

In my previous question, it was shown that $f(x,y)$ is $M\times M$ measurable. I am trying to show now that $f$ is $m\times m$ summable.

Is my approach correct?

Note that when $y<x$ we have that $x^2+y^2\leq 2x^2$ and $x^2-y^2\geq 0$. Then, $$\int_{0}^{1} |f_x| dm(y) \geq \int_{0}^{x} \frac{x^2-y^2}{(x^2+y^2)^2} dm(y)\geq \int_{0}^{x}\frac{x^2-y^2}{4x^4} dm(y)$$

Now, $$\int_{0}^{x} \frac{x^2-y^2}{4x^4} dm(y) = \frac{y}{4x^2}-\frac{y^3}{12x^4} \bigg|_{y = 0}^{y=x} = \frac{1}{6x}\rightarrow \infty$$

as $x\rightarrow 0$. Hence our function is not $m\times m$ summable.


The last part of your proof isn't right. Suppose you had obtained

$$\int_0^1|f(x,y)| \,dy \ge \frac{1}{\sqrt x}.$$

Then $1/\sqrt x \to \infty,$ but that doesn't show lack of summability. All we would know is that

$$\int_0^1\int_0^1 |f(x,y)|\,dx\, dy \ge \int_0^1 \frac{1}{\sqrt x} \, dx = 2,$$

which doesn't tell you much.

However, you obtained

$$\int_0^1|f(x,y)| \,dy \ge \frac{1}{6 x}.$$


$$\int_0^1\int_0^1 |f(x,y)|\, dy \,dx\ge \int_0^1 \frac{1}{6x} \, dx = \infty.$$

So it's not $1/(6x)\to \infty$ that is crucial, it's the lack of summability of $1/(6x)$ on $[0,1]$ that gives the desired conclusion.


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