This is a theorem in Yeh's Real Analysis :
Let $(X, \mathbf{A}, \mu)$ and $(Y, \mathbf{B}, v)$ be two finite measure spaces. Consider the
product measure space $(X \times Y, \sigma(\mathbf{\mathbf { A }} \times \mathbf{B}), \mu \times v) .$ Let $\mathbf{P}$ be the collection of nonnegative
functions on $X \times Y$ of the type $$\psi(x, y)=\sum_{j=1}^{k} c_{j} \mathbf{1}_{A_{j}}(x) \mathbf{1}_{B_{j}}(y)$$ where $c_{j}>0, A_{j} \in \mathbf{A},$ and $B_{j} \in \mathbf{B}$ for $j=1, \ldots, k .$ Then forevery nonnegative extended
real-valued $\sigma(\mathbf{A} \times \mathbf{B})$ -measurable function $f$ on $X \times Y,$ there exists an increasing sequence
$\left(\psi_{n}: n \in \mathbb{N}\right)$ in $\mathbf{P}$ such that $\psi_{n} \uparrow f$ a.e. on $X \times Y .$
The suthor first approximate $f$ using simple functions in the product measure space, and then approximate sets in $\sigma(\mathbf{\mathbf { A }} \times \mathbf{B})$ by sets in $\mathbf{\mathbf { A }} \times \mathbf{B}$.
However, I find that the proof does not completely solve the problem. Instead, it only proves that there exists a sequence of $\psi_{n}$ that converges a.e. to $f$, but they may not be monotone.
Can anyone give a correct proof to this theorem please ?
Best Answer
This seems to be wrong.
Consider $X=Y=[0,1]$ with the Borel $\sigma$-algebra and the Lebesgue measure. It is possible to show that there exists a Borel measurable set $A \subseteq [0,1]^2$ which has strictly positive Lebesgue measure and contains no rectangle of positive Lebesgue measure, i.e. if $I,J \subseteq [0,1]$ are Borel sets with $I \times J \subseteq A$ then $\lambda^1(I)=0$ or $\lambda^1(J)=0$ (see this question). Now consider the indicator function $f:=1_A$. Let $g$ be a simple function of "product type" $$g(x,y) = \sum_{k=1}^n c_k 1_{I_k}(x) 1_{J_k}(y)$$ such that $0 \leq g \leq f$. Without loss of generality, $c_k>0$ for all $k$. Since $g$ is strictly positive on $I_k \times J_k$, it follows that $f|_{I_k \times J_k}=1$, and so $$I_k \times J_k \subseteq A.$$ By our choice of $A$, this means that $\lambda^1(I_k)=0$ or $\lambda^1(J_k)=0$. As $k$ is arbitrary, this gives $g=0$ Lebesgue almost everywhere. Since $\lambda^2(\{f=1\})=\lambda^2(A)>0$, this shows that there cannot exist a sequence of simple functions $(g_k)_{k \geq 1} \subseteq \mathbf{P}$ with $g_k \uparrow f$ almost everywhere.