Prove or disprove: $(I^2,\mathcal{B},\mu+\nu)$ is localizable. Definition 1 If $(X_i,\mathcal{S}_i,\mu_i)$ are measure spaces for all $i$ in some index set $I$, where the sets $X_i$ are disjoint, the direct sum of these measure spaces is defined by taking $X=\bigcup_iX_i$, letting $\mathcal{S}:=\{A\subset X:A\cap X_i\in\mathcal{S}_i\;\mathrm{for}\;\mathrm{all}\;i\}$, and $\mu(A):=\sum_i\mu_i(A\cap X_i)$ for each $A\in\mathcal{S}$.
Definition 2 A measure space is called localizable iff it can be written as a direct sum of finite measure spaces.
Proposition
(a) Any $\sigma$-finite measure space is localizable.
(b) Any direct sum of $\sigma$-finite measure spaces is localizable.
Problem
Consider the unit square $I^2$ with Borel $\sigma$-algebra. For each $x\in I:=[0,1]$ let $I_x$ be the vertical interval $\{(x,y):0\leq y\leq1\}$. Let $\mu$ be the measure on $I^2$ given by the direct sum of the one-dimensional Lebesgue measures on each $I_x$. Likewise, let $J_y:=\{(x,y):0\leq x\leq1\}$ and let $\nu$ be the measure on $I^2$ given by the direct sum of the one-dimensional Lebesgue measures on each $J_y$. Let $\mathcal{B}$ be the collection of sets measurable for both direct sums $\mu$ and $\nu$ in $I^2$. Prove or disprove: $(I^2,\mathcal{B},\mu+\nu)$ is localizable.
Hint: Assuming the continuum hypothesis, the following problem applies to Lebesgue measure:
Product measure on two uncountable well-ordered sets
My question:
It seems to me that the problem given in the hint can be used as a counterexample. But I don't understand how that problem is related to this problem via continuum hypothesis.
 A: How is the Continuum Hypothesis equivalent to the existence of a well-ordering on $\Bbb R$ whose bounded initial segments are countable?
On a problem concerning surface measurable sets
According to a known theorem on the plain measurable sets $L$, a plain surface measurable set is measurable (linearly) on almost any straight line of any parallel beam. M. Zalcwasser recently gave an elementary proof of this theorem and posed the problem if the converse of the theorem in question is also true. We will show using the theorem of M. Zermelo that the answer is negative.
Theorem There exists a plane set which is of zero measure on any straight line, but which is not surface measurable.
Demonstration. For all the firm planes sets of positive surface measure forming a power set $P$ of the continuum, there exists (according to the theorem of M. Zermelo) a well ordered set
\begin{equation}
F_1,F_2,F_3,\ldots,F_\omega,F_{\omega+1},\ldots,F_\alpha,\ldots\quad(\alpha<\Omega_0)\quad(1)
\end{equation}
of the type $\Omega_0$ (or $\Omega_0$ is the smallest transfinite number corresponding to the power of the continuum) consisting of all sets $F$ which belong to $P$.
Similarly, for the set of all the points of the plane forming a power set of the continuum, there exists a transfinite sequence of the type $\Omega_0$
\begin{equation}
p_1,p_2,p_3,\ldots,p_\omega,p_{\omega+1},\ldots,p_\alpha,\ldots\quad(\alpha<\Omega_0)\quad(2)
\end{equation}
formed from all points $p$ of the plane.
We will now rely on the following
Lemma Any firm plane set which is of zero measure on any line parallel to a given line, is of zero surface measure.
The proof of this lemma (using M. Borel's theorem) does not offer any difficulty.
Corollary For any firm plane set $F$ with a positive surface measure and for any given line $D_0$, there exists a line $D$ parallel to $D_0$ and such that the linear measure of the set $FD$ is positive.
Indeed, $F$ being firm, the set $FD$ is (for any line $D$) firm. If $FD$ were of zero measure for any line $D$ parallel to $D_0$, the set $F$ would be, according to our lemma, of zero surface measure, contrary to the hypothesis. There is therefore a line $D$ parallel to $D_0$ and such that the set $FD$ is not of zero measurement. Therefore, $FD$ being firm, it is a set of positive measurement. This concludes the demonstration.
This corollary established, let $ q_1 $ be the first term of sequence (2) which is a point of the set $ F_1 $. Let $ \alpha $ be a given index, $ 1 <\alpha <\Omega_0 $, and assume that we have already defined all the points $ q_\xi $ with $ \xi <\alpha $. Let $G_\alpha$ denote the set of all lines $q_\mu q_\nu$ with $\mu<\nu<\alpha$. Like $\alpha <\Omega_0$, we can easily conclude that the set of all the lines that belong to $G_\alpha$ has a power less than that of the continuum. It follows from there the existence of a line $D_0$ which is not parallel to any of the lines forming $G_\alpha$. The set $F_\alpha$ being of positive surface measure, there exists, according to the corollary, a line $D$ parallel to $D_0$ and such that the linear measure of the set $F_\alpha D$ is positive. Line $D$, as parallel to $D_0$, does not coincide with any of the lines forming $G_\alpha$. The set of these lines being of lower power than that of the continuum, it follows that the points of intersection of the line $D$ with the lines of $G_\alpha$ form a set of lower power than that of the continuum. The set $F_\alpha D$ being of positive linear measure, therefore of power of the continuum, there exists a point $p$ of $F_\alpha$ which does not belong to any of the lines forming $G_\alpha$.
We have thus established the existence of a point $p$ in the set $F_\alpha$ which does not lie on any of the lines $q_\mu q_\nu$ with $\mu <\nu <\alpha$. Let $q_\alpha$ be the first term of the sequence (2) enjoying this property of $p$.
The transfinite sequence of points
\begin{equation}
q_1,q_2,q_3,\ldots,q_\omega,q_{\omega+1},\ldots,q_\alpha,\ldots\quad(\alpha<\Omega_0)\quad(3)
\end{equation}
is thus defined by the transfinite induction. Let $E$ be the set of all the points of this sequence. It follows immediately from the definition of the sequence (3) that any set $F_\alpha$ with $\alpha> \Omega_0$ admits at least one point in common with the set $E$ and that no three points of $E$ are located on a line.
Any straight line on the plane therefore has at most two points in common with the set $E$. Now, $E$ having at least one point in common with any firm set of positive surface measure, the complement of $E$ cannot be of positive surface measurement (since any set of positive surface measure contains a firm subset of positive surface measure). The set $E$ cannot therefore be of zero surface measure. However, $E$ cannot be of positive surface measure either, because $E$ would then contain a firm subset $F$ of positive surface measure and there would consequently exist a line $D$ on which the linear measure of $F$ would be positive; but it is impossible, since $D$ intersects with $E$, and therefore with $F$, at two points at most. We conclude that the set $E$ is not surface measurable.
We have thus shown (using M. Zermelo's theorem) that there exists a plane set $E$ having at most two points in common with any line and which is not surface measurable, - that is more than it was necessary to demonstrate.
Note that an analogous theorem for M. Borel's measure would be trivial: indeed, any non-measurable set $L$ (linearly) formed by points of a circumference is a non-measurable plane set $B$, having at most two points in common with any straight line.
Note also that one can easily deduce from the theorem proof the existence of an univariate real-valued function whose geometric image is not surface measurable.
Indeed, the set $E$ admits at most two common points with any line parallel to the ordinate axis. Let us consider those which have exactly two points in common with $E$. Let us choose on each of these lines the one of the two points of $E$ which is higher; let $E_1$ be the set of points thus obtained (it can also be an empty set). Let $E=E_1+E_2$. The set $E$ being not superficially measurable, we conclude that at least one of the sets $E_1$ and $E_2$ is not superficially measurable; let us designate it by $G$. It is obvious that $G$ admits at most one point in common with any line parallel to the ordinate axis (since it was thus for $E_1$ and $E_2$).
Let us now define the function $f(x)$ of a real variable $x$ as follows. Let $x_0$ be a real number given. The line $x = x_0$ has at most one point in common with the set $G$; if such a point exists and $y_0$ is its ordinate, let $f(x_0) = y_0$, otherwise let $f(x) = 0$.
The function $f(x)$ is thus defined for all the real values ​​of $x$. The geometric image of the function $f(x)$ is obviously the set $G + H$ where $H$ is a set of points on the abscissa axis, therefore a set of zero surface measure; Since $G$ is not surface measurable, so is $G + H$. This concludes the demonstration.
