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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.

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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.

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