Tonelli and Fubini for the completed product, with only almost-everywhere section measurability

By [L1], choose a product-measurable function $g$ such that $f=g$ almost everywhere for $\overline{\mu \times \nu }$. Let $N:=\{(x,y):f(x,y)\mathrm{\ne }g(x,y)\}$, so $\overline{\mu \times \nu }(N)=0$. By the completion definition, choose a product-measurable null set $Z$ with $N\subseteq Z$.

Apply the nonnegative case of [L2] to ${\mathbf{1}}_Z$. Since [L2] gives $(\mu \times \nu )(Z)=0$, Tonelli yields ${\int }_X\nu (Z_x)d\mu =0,{\int }_Y\mu (Z^y)d\nu =0.$ Hence $\nu (Z_x)=0$ for $\mu$-almost every $x$ and $\mu (Z^y)=0$ for $\nu$-almost every $y$. Because $N_x\subseteq Z_x$ and $N^y\subseteq Z^y$, the equalities $f_x=g_x$ and $f^y=g^y$ fail only on null sections of the completed factor spaces. For such $x$, the section $f_x$ is almost everywhere equal to the measurable section $g_x$, so [L4] makes $f_x$ $\bar{\mathcal{B}}$-measurable; similarly for $f^y$.

In the nonnegative case, apply Tonelli from [L2] to $g$. Since $f=g$ almost everywhere on the complete product space, the completed integral of $f$ equals that of $g$, and the section integrals agree for the almost-everywhere parameters isolated in step 1.2. This proves part 1.

If $f\in L^1(\overline{\mu \times \nu })$, apply [L1] separately to $\mathrm{Re}f$ and $\mathrm{Im}f$. This gives product-measurable real-valued functions $u,v$ such that $u=\mathrm{Re}f$ and $v=\mathrm{Im}f$ almost everywhere. Put $g:=u+iv$. Then $g$ is product-measurable, $g=f$ almost everywhere, and [L3] applied to $\mathrm{\mid }f\mathrm{\mid }$ and $\mathrm{\mid }g\mathrm{\mid }$ shows $g\in L^1(\mu \times \nu )$. The $L^1$ case of [L2] applies to $g$, step 1.2 transfers the almost-everywhere section measurability from $g$ to $f$, and [L3] transfers the equality of integrals from $g$ to $f$. This proves part 2.