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The function $(x-y)/(x+y{)}^3$ on the unit square is not Lebesgue integrable. On $0<y<x/2$, its absolute value is at least $4/(27x^2)$, whose area integral diverges logarithmically. Its two signed iterated integrals nevertheless exist, ignoring null boundary sections, and equal $1/2$ and $-1/2$. This illustrates why absolute integrability is required for Fubini's theorem.

The domain of integration is the set $D$ over which an integral is taken. For a multiple integral, inequalities or parameter ranges specify which points belong to $D$; a change of variables transforms both these bounds and the integration element through its Jacobian determinant. For example, a polar region $c\le r\le 1+\cos \varphi$ exists only at angles where $1+\cos \varphi \ge c$.

The set over which an integral is taken. For a double integral, boundary inequalities describe the allowed coordinate pairs, and slices of this set determine the limits in an iterated integral. The domain need not be a rectangle or connected. A change of variables transforms both the integrand and this set.

Define the displayed function away from the origin and set $f(0,0)=0$. Every section in either coordinate is integrable on $[-1,1]$ and is an odd function, so both orders of iterated integrals on $[-1,1{]}^2$ are zero. However, on the annulus $\epsilon <\sqrt{x^2+y^2}<1$, polar integration givesThus the function is not Lebesgue integrable jointly. Reflection interchanges its positive part and negative part, whose integrals are both infinite. Existence and even agreement of finite iterated integrals do not establish the absolute integrability hypothesis of Fubini's theorem.

An iterated integral performs successive one-variable integrals. In the displayed expression, $y$ is held fixed during the inner integration over $x$, and the result is then integrated over $y$. The inner bounds may depend on that fixed outer variable. By contrast, an inner bound depending on an already integrated variable does not specify such a nested operation without further interpretation. For example, the triangle $0\le y\le 1$, $2y\le x\le 2$ gives ${\int }_0^1{\int }_{2y}^{2}f(x,y)dxdy$, or, in the other order, ${\int }_0^2{\int }_0^{x/2}f(x,y)dydx$. Fubini's theorem and Tonelli's theorem supply conditions under which these iterated expressions equal the same multiple integral.

The sequence in which variables are integrated in an iterated integral. Reversing the order changes the slicing of the domain of integration, often requiring more than one integral. Fubini's theorem permits interchange for absolutely integrable functions, while Tonelli's theorem permits nonnegative measurable integrands even when the integral is infinite. Conditional convergence requires additional care; geometric reversal alone is not a general justification for interchanging improper integrals.

For $-1\le y\le 0$, a horizontal slice runs from $x=-1$ to the line $x=y$. For $0\le y\le 1$, it runs from $x=-1$ to the negative branch $x=-\sqrt{y}$ of the parabola. Thus the reversed iterated integral isThe positive branch $x=\sqrt{y}$ is outside the original integration region.

The stated $x$ bounds depend on $y$, so $y$ must be held fixed while the integral over $x$ is performed. The intended iterated integral isThe original PDF places the two sets of bounds beside the opposite integral signs relative to the printed $dxdy$. That placement cannot be read literally as a conventional nested integral; the dependence of the lower $x$ bound on $y$, and the explicit instruction to integrate over $x$ first, specify the intended region unambiguously.

Fubini's theorem for the product of two sigma-finite measure spaces states that an integrable measurable function on the product has integrable sections almost everywhere, the section integrals define Lebesgue integrable functions, and both iterated integrals equal the product integral. The integrability hypothesis is absolute integrability. For nonnegative measurable functions, Tonelli's theorem gives equality of the possibly infinite product and iterated integrals without first assuming finiteness.

The displayed function is not Lebesgue integrable. Its value at the origin can be assigned arbitrarily, since a point is a null set. On $0<x<1$ and $0<y<x/2$,Consequently Tonelli's theorem gives the divergent lower boundThe signed iterated integrals illustrate why the hypothesis matters. For $x>0$, $f(x,y)={\partial }_y[y/(x+y{)}^2]$, so integration first in $y$ and then in $x$ gives $1/2$. Integrating first in $x$ gives $-1/2$, using $f(x,y)=-{\partial }_x[x/(x+y{)}^2]$. Exceptional boundary sections have measure zero. Thus finite iterated values do not establish product integrability.