Approximating a product-measurable function from below On page 198 of Dunford and Schwartz, Vol. I, in the proof of part (b) of Lemma III.11.16, the following assertion is made without proof (or reference).  Let $(X,\mathscr{X},\mu)$ and $(Y,\mathscr{Y},\nu)$ be finite measure spaces, and $(X\times Y, \mathscr{X}\otimes\mathscr{Y},\mu\times\nu)$ be their product space.  If $f\colon X\times Y\to [0,\infty)$ is measurable with respect to the product $\sigma$-algebra $\mathscr{X}\otimes\mathscr{Y}$, then $f$ can be approximated a.e. by a sequence $f_n$ of finite linear combinations of characteristic functions of sets of the form $A\times B$, $A\in\mathscr{X}$, $B\in\mathscr{Y}$ (this is standard so far), and one may in addition assume that $f_n\leqslant f $ a.e. for all $n$.  Is this (last statement) entirely obvious?
 A: It turns out that the statement is wrong.  As I was browsing through the questions on the right of this screen, I came across the books by Fremlin and Bogachev (in the question with title "Uniqueness of product measure (non $\sigma$-finite case)").  Exercise 3.10.62 of Bogachev's book is as follows (p. 235).  Given atomless probability spaces $(X_1,\mathcal{A_1},\mu_1)$ and $(X_2,\mathcal{A_2},\mu_2)$, there exists a set $A\in\mathcal{A_1}\otimes\mathcal{A_2}$, with $(\mu_1\otimes\mu_2)(A)>0$, such that $(\mu_1\otimes\mu_2)((A_1\times A_2)\setminus A)>0$ for any sets $A_1\in\mathcal{A}_1$, $A_2\in\mathcal{A_2}$ with $\mu_1(A_1)\mu_2(A_2)>0$.  
In fact this is proved in a paper by Erdos and Oxtoby (Trans. AMS 79 (1955), 91--102).  Let $(X_1,\mathcal{A_1},\mu_1)$ and $(X_2,\mathcal{A_2},\mu_2)$ be both the unit interval $[0,1]$ with Lebesgue measure.  Let $D'$ be a fat Cantor set in the unit interval, i.e., a nowhere dense compact set of Lebesgue measure $\lambda_1(D')\in (0,1)$, where $\lambda_1$ is linear Lebesgue measure.  Let $D:=D'\cup(-D')$.  Now set $A:=\{ (x,y)\in [0,1]^2\colon x-y\in D\}$.  This is Borel measurable, being the inverse image of $D$ under the continuous map $(x,y)\mapsto x-y$.  With $\lambda_2$ denoting two-dimensional Lebesgue measure, for $A_1,A_2$ Borel subsets of $[0,1]$ one then has that \begin{align*}\lambda_2(A^c\cap (A_1\times A_2))&=\int\int\mathbf{1}_{D^c}(x-y)\mathbf{1}_{A_1}(x)\mathbf{1}_{A_2}(y)\, dx\, dy\\ &= \int_0^1\int_{-1}^1\mathbf{1}_{D^c}(u)\mathbf{1}_{A_1}(u+v)\mathbf{1}_{A_2}(v)\, du\, dv = \int_{[-1,1]\cap D^c} f(u)\, du,\end{align*} where $f(u):=\int\mathbf{1}_{A_1}(u+v)\mathbf{1}_{A_2}(v)\, dv =\lambda_1(A_1\cap(A_2+u))$.  The function $f$ is non-negative and continuous, $f(-u)=\mathbf{1}_{-A_1}\ast\mathbf{1}_{A_2}(u)$ being the convolution of two $L^2$-functions; moreover, the same calculation shows that $\int f(u)\, du=\lambda_2(A_1\times A_2)$, which is positive if $\lambda_1(A_1)\lambda_1(A_2)>0$.  It follows that $f$ is positive on some interval $I=(a,b)$, say, contained in $[-1,1]$ (since $f=0$ outside $[-1,1]$).  Since $I\cap D^c$ is open and non-empty (as $I$ cannot be contained in $D$ it being nowhere dense in $[-1,1]$), it has positive linear Lebesgue measure.  Thus \begin{equation*}\lambda_2(A^c\cap (A_1\times A_2))=\int_{[-1,1]\cap D^c} f(u)\, du\geqslant \int_{I\cap D^c} f(u)\, du>0\end{equation*} when $\lambda_1(A_1)\lambda_1(A_2)>0$.  Note also that, by the same calculation as above, one has that \begin{equation*} \lambda_2(A)=\lambda_2(A\cap [0,1]^2)= \int_D \lambda_1([0,1]\cap ([0,1]+u))\, du=\int_D (1-|u|)\, du\end{equation*} which is positive, since $D$ is of positive linear Lebesgue measure.
