= Solution
{c}
For <sigma-finite measures>, <Tonelli's theorem> states that the double <integral> of a nonnegative measurable function equals either iterated integral, allowing $+\infty$. <Fubini's theorem> states that if a measurable function is absolutely integrable, its sections are integrable almost everywhere, both iterated integrals exist, and they equal the double integral.
Assign any value at $(0,0)$, which does not affect <Lebesgue integrability>. Near that corner set $s=x+y$ and $t=(x-y)/(x+y)$. Then $x=s(1+t)/2$, $y=s(1-t)/2$, and the absolute <Jacobian determinant> is $s/2$. On $0<s<1$, the full range is $-1<t<1$, so
$$
\iint_{x+y<1}|F(x,y)|\,dx\,dy
=\frac12\int_0^1 s^{2-\alpha}\,ds\int_{-1}^1|t|\,dt
=\frac12\int_0^1s^{2-\alpha}\,ds.
$$
The remaining part of the unit square has $x+y\geq1$ and is harmless. The <improper power integral> is finite precisely when $\alpha<3$. Under the given restriction,
$$
\boxed{F\text{ is Lebesgue integrable exactly for }2<\alpha<3.}
$$
For those values, the change $(x,y)\mapsto(y,x)$ preserves the square and changes the sign of $F$. Absolute integrability justifies this change of variables, giving
$$
\boxed{\iint_{\mathbb R^2}F(x,y)\,dx\,dy=0.}
$$
For $\alpha\geq3$ the positive and negative parts each diverge, so symmetry does not define a Lebesgue integral.
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