counter-example of Fubini’s theorem for the Lebesgue integral

The following observation demonstrates the necessity of theintegrability assumption in Fubini’s theorem. Let

$$Q=\{(x,y)\in {ℝ}^2:x\ge 0,y\ge 0\}$$

denote the upper, right quadrant.Let $R\subset Q$ be the region in the quadrant bounded by the lines$y=x,y=x-1$, and let let $S\subset Q$ be a similar region, but thistime bounded by the lines $y=x-1,y=x-2$. Let

$$f={\chi }_S-{\chi }_R,$$

where $\chi$ denotes a characteristic function.

Observe that the Lebesgue measure of $R$ and of $S$ is infinite.Hence, $f$ is not a Lebesgue-integrable function. However for every$x\ge 0$ the function

$$g(x)={\int }_0^{\mathrm{\infty }}f(x,y)𝑑y$$

is integrable.Indeed,

Similarly, for $y\ge 0$, the function

$$h(y)={\int }_0^{\mathrm{\infty }}f(x,y)𝑑x$$

is integrable. Indeed,

$$h(y)=0,y\ge 0.$$

Hence, the values of the iterated integrals

$${\int }_0^{\mathrm{\infty }}g(x)𝑑x=-1,$$

$${\int }_0^{\mathrm{\infty }}h(y)𝑑y=0,$$

are finite, but do not agree. This does not contradict Fubini’stheorem because the value of the planar Lebesgue integral

$${\int }_{Q}f(x,y)𝑑\mu (x,y),$$

where $\mu (x,y)$ is the planar Lebesgue measure, is not defined.