Use of Fubini's theorem to check if equalities holds

Question

I'm working on the following problem.

Would you check if my solution makes sense? I feel like I need more details in my argument. I don't know if it suffices.

Problem:

Let $D = \{(x, y) \in \mathbb{R}^2 \mid -1 \leq x, y \leq 1\}$, and let $f : D \rightarrow \mathbb{R}$ be the following function:

$$f(x, y) = \begin{cases} \frac{x^5-y^5}{x^2+y^2} & \text{if } x \neq 0 \text{ or } y \neq 0 \\ +\infty & \text{if } x = y = 0 \end{cases}$$

Do the following equalities hold? $$\int_D f d\lambda^2 = \int_{-1}^1 \int_{-1}^1 f(x,y) d\lambda(x) d\lambda(y) = \int_{-1}^1 \int_{-1}^1 f(x,y) d\lambda(y) d\lambda(x)$$

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My attempt:

Use Fubini's theorem: if a function is integrable over a product space, then its iterated integrals are also integrable and yield the same result as the full integral.

In this case, since the function is integrable over $D$, Fubini's theorem applies and we can interchange the order of integration without changing the value of the integral.

Therefore, we would get:

$\int_D f d\lambda^2 $ to be equal to the double integral from $-1$ to $1$ over $f(x,y)\lambda(x)\lambda(y)$ which would be equal to the double integral from $-1$ to $1$ over $f(x,y)\lambda(y)\lambda(x)$

Answer 1

The bulk of the work here is in proving that $f$ is integrable over the product space! Concretely, this amounts to proving that $$ \int_D |f| d\lambda^2 < \infty.$$ You haven't done this.

My suggestion would be to redefine the value of $f$ $(0, 0)$ to be $0$, rather than $\infty$. Since $\{ (0, 0) \}$ is a set of measure zero, redefining $f$ in this way won't affect our conclusions.

My recommended next steps would be to:

• Show that $f$ is continuous on $D$, including at the point $(0, 0)$.
• Deduce that $f$ is bounded on $D$. (Think about what you know about the image of a continuous function on a compact space.)
• Deduce that $\int_D |f| d\lambda^2 < \infty$.