On a product of $\sigma$-finite spaces, if one iterated integral of $\abs{f}$ is finite, then $f\in L^1$ and both iterated integrals of $f$ equal $\int f$.Suggest a slogan
Let \((X,\mcm,\mu)\) and \((Y,\mcn,\nu)\) be \(\sigma\)-finite measure spaces, and let \(f\colon X\times Y\to\CC\) be \(\mcm\otimes\mcn\)-measurable. Suppose that one of the iterated integrals of \(\abs{f}\) is finite: \[ \int_X\qty{\int_Y\abs{f(x,y)}\,d\nu(y)}d\mu(x)<\infty\quad\text{or}\quad\int_Y\qty{\int_X\abs{f(x,y)}\,d\mu(x)}d\nu(y)<\infty. \] Then \(f\in L^1(X\times Y,\mu\times\nu)\), and \[ \int_{X\times Y}f\,d(\mu\times\nu)=\int_X\qty{\int_Y f(x,y)\,d\nu(y)}d\mu(x)=\int_Y\qty{\int_X f(x,y)\,d\mu(x)}d\nu(y), \] where the inner integrals exist for almost every value of the outer variable.