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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.

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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 $\epsilon'\to 0$.

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