More specifically, consider the following particular situation: Let $I=[0,1]$ with the standard Borel $\sigma$-algebra. Consider functions $y:I\times I\to I$. Say that $y$ is *scalarly* measurable iff $y(\cdot,s)$ is measurable for every fixed $s\in I$. Say that $y$ is *simple* iff there is a partition of $I$ into finitely many disjoint measurable sets $A\_1,\ldots\kern.7mm A\_k$ such that $A\_i\owns t\mapsto y(t,\cdot)\in I^I$ is a constant function for all $i=1,\ldots\kern.7mm k$. Finally say that $y$ is *simply measurable* iff there is a sequence $\langle\kern.7mm y\_i:i\in\mathbb N\_0\kern.4mm\rangle$ with $y\_i$ simple for all $i\in\mathbb N\_0$ and such that $y\_i(t,s)\to y(t,s)$ for all $t,s\in I$ as $i\to\infty$. Trivially every simply measurable $y$ is scalarly measurable, and the question now is



>  If $y$ is scalarly measurable, is it also simply measurable? 



Is scalarly measurable simply measurable?
Since the total cardinality of the set of sequences of finite Borel partitions is continuum, you might just as well ask if there is a universal pointwise approximation scheme for all Borel functions on $[0,1]$ by simple functions. The answer is "No". Suppose that $P\_k$ is any fixed family of Borel partitions. WLOG $P\_{k+1}$ is a refinement of $P\_k$. Replacing each partition element by an appropriately chosen compact subset, we get a Cantor-type set with a tree-like sequence of partitions, so, assuming that the partitions separate points (otherwise everything is trivial), doing partitions into many pieces in finitely many steps during each of which we partition into 2 pieces, and cutting off finite branches (countably many points), we arrive at the question whether every Borel function $f$ on $\{0,1\}^\mathbb N$ is a pointwise limit of a sequence of functions $f\_k$ depending on first $k$ coordinates. The characteristic function of the set of sequences with infinitely many ones is the classical counterexample. Indeed, denote by $0\_{k\_1}1\_{\ell\_1}0\_{k\_2}\dots$ the (finite or infinite) sequence starting with $k\_1$ zeroes followed by $\ell\_1$ ones followed by... Suppose that $f\_k(0\_k)>\frac 13$ for all $k$. Then $0\_\infty$ gives us trouble, so $f\_{k\_1}(0\_{k\_1})\le \frac 13$. Now if $f\_{k\_1+\ell}(0\_{k\_1}1\_{\ell})<\frac 23$ for all $\ell$, then $0\_{k\_1}1\_\infty$ is a disaster. Thus $f\_{k\_1+\ell\_1}(0\_{k\_1}1\_{\ell\_1})\ge \frac 23$. Now try to convince the sequence to go below $\frac 13$ again feeding it with $0$, etc. If it gets stubborn at some point, it will fail to converge to the right limit. If it follows your lead eventually every time, it will fail to converge at all. 


This argument (after a few appropriate modifications) allows you to show that all Baire classes (pointwise limits of continuous functions, pointwise limits of pointwise limits, etc.) are different. My first impulse was to redirect you to AoPS, to post a small hint, and to vote to close but then I thought a bit about how "well" we teach our graduate students the elementary measure theory and decided to post a (reasonably) full answer. Feel free to ask any questions if something is unclear :-).


