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
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 $\Omega_0$
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
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.