Is scalarly measurable simply measurable?

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?

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.
• @ fedja: Thanks for your answer. From it I understand that your answer to my question is "no". However, I do not understand at all what precisely might be the formal deduction sequence you are sketching. Several questions arise. First I would like to know what is "universal pointwise approximation scheme for all Borel functions on $[0,1]$ by simple functions", and why existence of such is equivalent(?) to my question? Is there any reference where these things are explained? – TaQ Sep 15 '13 at 21:43
• If "universal pointwise approximation scheme ..." means a function $S$ associating with every set $B$ of Borel functions $I\to I$ a function $S(B):B\to\big(I^I\big)^{\mathbb N_0}$ such that $S(B)(x)(i)(t)\to x(t)$ as $i\to\infty$ for every fixed $t\in I$ and $x\in B$, and that is such that for every fixed $i\in\mathbb N_0$ there is a Borel partition $\mathcal B(B,i)$ such that $S(B)(x)(i)$ is simple and "over" this partition independently of $x\in B$, then I understand the existence of such a scheme $S$ to be rather trivially equivalent to my question. (cont.) – TaQ Sep 15 '13 at 22:59
• However, I still do not understand how the families $P_k$ possibly are related to such a scheme $S$. Is it possibly that $P_i=\{\kern.6mm\mathcal B(B,i):B\text{ set of Borel functions}\kern1mm\}$ ? – TaQ Sep 15 '13 at 22:59
• If this is what you have meant, and if the next step is to deduce that existence of $S$ implies that 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, and that this in turn has a classical counterexample, what is the purpose of mentioning that "this argument ... allows you to show that all Baire classes ... are different"? – TaQ Sep 15 '13 at 23:22
• @TaQ 1) You can list all Borel functions as $y(\cdot,s)$, so if you do it, you'll have to use one fixed sequence of partitions regardless of the function (that is what I meant by "universality"). 2) The main purpose of mentioning that "this argument ... different" was that some good textbooks I know do not bother to prove the statement about binary trees I used separately but discuss it in the context of the Baire hierarchy, so if you want to look it up there, that's the section you might want to go to. Also that's where I would do it in a measure theory course. :-). – fedja Sep 16 '13 at 1:23