For sigma-finite measures, the two section-measure integrals of a measurable set agree

Fix $n,m\ge 1$. Let ${\mathcal{D}}_{n,m}$ be the family of measurable subsets $F\subseteq X_n\times Y_m$ such that ${\int }_{X_n}\nu (F_x)d\mu ={\int }_{Y_m}\mu (F^y)d\nu \mathrm{.}$ If $F=(A\times B)\cap (X_n\times Y_m)$, then $F_x=(B\cap Y_m)$ for $x\in A\cap X_n$ and $\mathrm{\varnothing }$ otherwise, so both integrals equal $\mu (A\cap X_n)\nu (B\cap Y_m)$. Finite additivity gives the same equality for the algebra of [L2].

If $F_k\uparrow F$ inside $X_n\times Y_m$, then [L1] and [L4] give ${\int }_{X_n}\nu ((F_k{)}_x)d\mu \uparrow {\int }_{X_n}\nu (F_x)d\mu$, and similarly on $Y_m$. Hence ${\mathcal{D}}_{n,m}$ is closed under increasing unions. For $F\in {\mathcal{D}}_{n,m}$, put $F^c=(X_n\times Y_m)\setminus F$. Each section-integral of $F^c$ equals the finite number $\mu (X_n)\nu (Y_m)$ minus the corresponding section-integral of $F$. This follows by adding the complementary sections and integrating; all quantities are finite because they are bounded by that finite product. Thus $F^c\in {\mathcal{D}}_{n,m}$. If $F_k\downarrow F$, their relative complements increase to $F^c$; closure under increasing unions and complements therefore gives $F\in {\mathcal{D}}_{n,m}$. The class is now proved monotone in both directions and contains the relative rectangle algebra by step 1.1. Applying [L2] and [L3] on $X_n\times Y_m$ yields every product-measurable subset of that rectangle.

Put $E_{n,m}:=E\cap (X_n\times Y_m)$. Step 2.1 gives ${\int }_{X_n}\nu ((E_{n,m}{)}_x)d\mu ={\int }_{Y_m}\mu ((E_{n,m}{)}^y)d\nu \mathrm{.}$ Now $(E_{n,m}{)}_x=E_x\cap Y_m$ for $x\in X_n$ and $\mathrm{\varnothing }$ otherwise, so as $n,m\to \mathrm{\infty }$ the two integrands increase pointwise to $\nu (E_x)$ and $\mu (E^y)$.

Applying [L4] on both sides of step 3.1 and then letting $n,m\to \mathrm{\infty }$ gives ${\int }_X\nu (E_x)d\mu ={\int }_Y\mu (E^y)d\nu \mathrm{.}$ This is the claimed equality.