interchanging integrals Why does $$\int_0^{y/2} \int_0^\infty e^{x-y} \ dy \ dx \neq \int_0^\infty \int_0^{y/2} e^{x-y} \ dx \ dy$$ The RHS is 1 and the LHS side is not. Would this still be a legitimate joint pdf even if Fubini's Theorem does not hold?

$$\int_0^{\infty} \int_0^{y/2} \exp(x-y) dx dy = \int_0^{\infty} \int_{2x}^{\infty} \exp(x-y) dy dx$$

Note that both, not surprisingly, yield the same answer.

$$\int_0^{\infty} \int_0^{y/2} \exp(x-y) dx dy = \int_0^{\infty} (\exp(-y/2) - \exp(-y)) dy = 1$$

$$\int_0^{\infty} \int_{2x}^{\infty} \exp(x-y) dy dx = \int_0^{\infty} \left. - \exp(x-y) \right|_{2x}^{\infty} dx = \int_0^{\infty} \exp(-x) dx = 1$$