# Showing Iterated Lebesgue Integrals Do Not Coincide

Let $$f(x,y) = \frac{x}{1 - y^2}$$ on $$(-1,1) \times (-1, 1)$$. I have two iterated Lebesgue integrals

$$I =\int_{-1}^1 \int_{-1}^1 f(x,y)dxdy ~~~~~~~~~~~\text{and}~~~~~~~~~~~ J =\int_{-1}^1 \int_{-1}^1 f(x,y)dydx,$$

and I would like to show that $$I \neq J$$.

For $$I$$:

I can fix a $$y \in (-1,1)$$, and then let $$g(x) = \frac{x}{1-y^2}$$, which is continuous and bounded on $$[-1,1]$$. Hence, $$g(x)$$ is Riemann integrable and is equivalent to the Lebesgue integral. Evaluating, $$\int_{-1}^1 g(x)dx = 0$$. Since this holds for any $$y \in (-1,1)$$, we have the Lebesgue integral of the zero function, which is clearly zero. Thus, $$I$$ exists and $$I = 0$$.

For $$J$$:

Surely $$J$$ is undefined and $$I \neq J$$. But fixing an $$x \in (-1,1)$$ does not give a bounded function, and so I cannot say that the Riemann and Lebesgue integrals coincide here necessarily as I did above. Not sure about other Theorems I could apply to show this?

Edit: I think it's enough to just apply the definition: $$\int_{-1}^1 f^{+}dy$$ and $$\int_{-1}^1 f^{-1}dy$$ are both $$\infty$$, which implies $$\int_{-1}^1 fdy = \infty$$, and so it follows that $$J$$ undefined.

The Riemann integrals $$\int_{-1+\epsilon}^{1-\epsilon'}f(x,y)\mathrm dy$$ exist for each $$\epsilon,\epsilon'>0$$. So they equal the Lebesgue integrals over the same domain. Thus they both run off to infinity as $$\epsilon\to 0$$ and $$\epsilon'\to 0$$.