Question regarding Fubini's theorem and a piecewise function

Question

Specifically, the question is:

Consider the function $f:[0,1]\times[0,1]\rightarrow\mathbb{R}$ defined by $$f(x,y)=\Bigg\{\begin{array}{cc} y^{-2}&0<x<y<1,\\ -x^{-2}&0<y<x<1,\\ 0&otherwise. \end{array}$$ Show by direct computation that $$\int_0^1\left(\int_0^1f(x,y)dx\right)dy\neq\int_0^1\left(\int_0^1f(x,y)dy\right)dx.$$ Why does this not contradict the Fubini's Theorem?

The first thing I need clarification on is how to evaluate these integrals.

Fubini's theorem has many hypotheses. If, for example, we were working with a non-continuous, non-measurable, or non-integrable function, then it wouldn't apply. I need assistance identifying the hypothesis that isn't met.

$\textit{Edit:}$ I removed my incorrect thoughts on this problem to reduce clutter.

Answer 1

$f$ is defined by 1/monomials. This should hint to you that 0 is not nice, and you should be able to easily show $f$ is not continuous at 0. Here's a graph of $f$ that clearly shows an issue at $0$,

and a graph of $|f|$,

Fubini holds for discontinuous functions as well as long as they are Lebesgue integrable, but you should also be able to show that this function is not.

To compute the iterated integral, you should compute the inner integral first. I'll do one in full:

$$LHS = \int_0^1 \left( \int_0^yy^{-2} dx - \int_y^1 x^{-2} dx\right) dy \\= \int_0^1 y^{-1} + \left.x^{-1}\right|_y^1 dy = \int_0^1 y^{-1} + 1 - y^{-1} dy = \int_0^1 dy = 1$$

For $RHS$, you should get $-1$.

Answer 2

For the left-hand-side: Fix some $y\in (0, 1)$. We have \begin{align*} \int_0^1 f(x,y) dx &= \int_0^y f(x,y) dx + \int_y^1 f(x,y) dx \\ &= y^{-2}x|_{x=0}^{x=y} + x^{-1} |_{x=y}^{x=1} \\ &= y^{-1} + (1 - y^{-1}) \\ &= 1. \end{align*} That is \begin{align*} \int_0^1 \left(\int_0^1 f(x,y) dx \right) dy = y|_{0}^1 = 1. \end{align*}

You might want to recompute the right-hand-side yourself. Also, you might want to reconsider that $f$ is integrable.