MATHEMATICS

I. N. BERNSTEIN

ON THE SET SWEPT OUT BY A LINE SEGMENT

(Presented by Academician I. G. Petrovskii, 4 IV 1962)

Let \(l(x)\) be a line in the plane \(x,y\), passing through the point \(x\) of the axis of abscissas and forming an angle \(\alpha(x)\ne 0\) with the positive direction of the \(x\)-axis. Denote by \(\Delta(x)\) the segment of the line \(l(x)\) between the axis of abscissas and the line \(y=1\). Suppose the line \(l(x)\) is defined for all \(x\) in the segment \([0,1]\). If \(\alpha(x)\) is a continuous function, then the segment \(\Delta(x)\), as the point \(x\) moves along the segment \([0,1]\), sweeps out an area not less than \(1/2\).

What is the situation if \(\alpha(x)\) is discontinuous?

In this note it will be shown that there exists such a measurable function \(\alpha(x)\) that \(\Delta(x)\) sweeps out zero area. Moreover, the function \(\alpha(x)\) can be constructed so that the endpoint of the segment \(\Delta(x)\), moving along the line \(y=1\), traverses the segment \(0 \le x \le 1\), hitting each point exactly once.

Let a parallelogram \(\pi\) be given. Fix two of its opposite sides (we shall call them bases).

We shall call an \(a\)-system a finite system of parallelograms located in \(\pi\), such that two opposite sides of each parallelogram lie on the bases of \(\pi\), the bases of \(\pi\) being completely covered by these sides, while the sides of different parallelograms lying on a base of \(\pi\) may intersect only at their endpoints. The set-theoretic sum of the parallelograms of a given \(a\)-system \(A\) will be denoted by \(|A|\).

Suppose that in some fixed parallelogram \(\pi\) an \(a\)-system \(A\) is given, consisting of parallelograms \(\pi_1,\ldots,\pi_n\), and suppose that in each parallelogram \(\pi_i\) \((i=1,\ldots,n)\) an \(a\)-system \(A_i\) is given (where as bases of \(\pi_i\) the sides lying on the bases of \(\pi\) are taken). Then all the parallelograms of all the systems \(A_i\) \((i=1,\ldots,n)\) form an \(a\)-system \(\widetilde A\) in \(\pi\). We shall say that \(\widetilde A\) is embedded in \(A\). If \(\widetilde A\) is embedded in \(A\), then, obviously, \(|\widetilde A|\subset |A|\) and \(\operatorname{mes}|\widetilde A|\le \operatorname{mes}|A|\).

Take a sequence of \(a\)-systems \(A^1,A^2,\ldots,A^n,\ldots\) in \(\pi\) such that \(A^n\) is embedded in \(A^{n-1}\) \((n=2,3,\ldots)\). Then \(|A^1|, |A^2|,\ldots, |A^n|,\ldots\) are nonempty closed sets, and moreover \(|A^{n-1}|\supset |A^n|\) \((n=2,3,\ldots)\). Hence their intersection
\[ B=\bigcap_{n=1}^{\infty}|A^n| \]
is a nonempty closed set, and
\[ \operatorname{mes} B=\lim_{n\to\infty}\operatorname{mes}|A^n|. \]

Such sets (intersections of sequences of embedded \(a\)-systems) will for brevity be called \(b\)-sets.

From the definition of a \(b\)-set it follows that for any \(b\)-set \(B\) and for any \(\varepsilon>0\) one can find an \(a\)-system \(A\) such that \(|A|\supset B\) and \(\operatorname{mes}|A|<\operatorname{mes} B+\varepsilon\).

Let \(\inf \operatorname{mes} B=\beta\), where the greatest lower bound is taken over all \(b\)-sets \(B\subset \pi\). We shall prove that the greatest lower bound is attained: there exists a \(b\)-set \(B_0\subset \pi\) whose measure is equal to \(\beta\).

Let an \(a\)-system \(A\) be given in a parallelogram \(\pi\), and let the parallelogram \(\pi'\) be obtained from \(\pi\) by an affine transformation \(\alpha\). Under the transformation \(\alpha\), the \(a\)-system \(A\) passes into some \(a\)-system \(A'\) in \(\pi'\). We shall say that \(A'\) in \(\pi'\) is induced by the \(a\)-system \(A\) in \(\pi\). Note that the ratios \(\operatorname{mes}|A|/\operatorname{mes}\pi\) and \(\operatorname{mes}|A'|/\operatorname{mes}\pi'\) are equal.

Let an \(a\)-system \(\widetilde A\) in \(\pi\) be embedded in an \(a\)-system \(A\) in \(\pi\), and suppose that all \(A_i\) in \(\pi_i\) (see the definition of embedded \(a\)-systems) are induced by one and the same \(a\)-system \(A^*\) in \(\pi\), for all \(i=1,2,\ldots,n\). For this case we introduce the special notation
\[ \widetilde A=A\cdot A^*. \]
If \(\widetilde A=A\cdot A^*\), then \(\operatorname{mes}|\widetilde A|\leqslant \operatorname{mes}|A^*|\). Indeed,
\[ \operatorname{mes}|\widetilde A|\leqslant \sum_i \operatorname{mes}|A_i|= \]
\[ =\sum_i \operatorname{mes}|A^*|\frac{\operatorname{mes}\pi_i}{\operatorname{mes}\pi}= \]
\[ =\operatorname{mes}|A^*|\sum_i \frac{\operatorname{mes}\pi_i}{\operatorname{mes}\pi} =\operatorname{mes}|A^*|. \]

Fig. 1

Now choose a sequence of \(b\)-sets \(B_n\subset \pi\) such that
\[ \lim_{n\to\infty}\operatorname{mes} B_n=\beta. \]
Let, further, for each \(n=1,2,\ldots\) the \(a\)-system \(A_n\) in \(\pi\) be such that
\[ \operatorname{mes}|A_n|<\operatorname{mes} B_n+\frac{1}{2^n}. \]
Since for every \(a\)-system \(A\) the set \(|A|\) is, obviously, a \(b\)-set, we have \(\operatorname{mes}|A_n|\geqslant \beta\). Consequently,
\[ \lim_{n\to\infty}\operatorname{mes}|A_n|=\beta. \]

We now construct a new sequence of \(A\)-systems:
\[ \widetilde A_1=A_1,\qquad \widetilde A_n=\widetilde A_{n-1}\cdot A_n\quad (n=2,3,\ldots). \]
By what was proved above,
\[ \operatorname{mes}|\widetilde A_n|\leqslant \operatorname{mes}|A_n|. \]
Consequently,
\[ \lim_{n\to\infty}\operatorname{mes}|\widetilde A_n|=\beta. \]

Since \(\{\widetilde A_n\}\) is a sequence of embedded \(a\)-systems, it follows that
\[ B_0=\bigcap_{n=1}^{\infty}|\widetilde A_n| \]
is a \(b\)-set, and \(\operatorname{mes}B_0=\beta\). We now show that \(\beta=0\). Suppose the contrary: \(\beta>0\). We shall be able to construct a \(b\)-set \(B^*\), whose measure is less than that of \(B_0\), and thus arrive at a contradiction.

Thus, let \(\beta>0\). Then in \(B_0\) there is a density point \(P\) which does not lie on the base of \(\pi\). Consider the \(a\)-system \(A^*\) in \(\pi\) shown in Fig. 1. \(A^*\) consists of the four parallelograms \(\pi_1,\pi_2,\pi_3,\pi_4\). The parallelograms \(\pi_1\) and \(\pi_2\) are chosen so that their lateral sides, respectively, intersect on the straight line parallel to the base of \(\pi\) and passing through the point \(P\). Then, under the affine transformations carrying \(\pi\) respectively into \(\pi_1\) and into \(\pi_2\), the images of the point \(P\) coincide.

Let, under the affine transformation \(\alpha_i\) carrying \(\pi\) into \(\pi_i\) \((i=1,2,3,4)\), the \(b\)-set \(B_0\) pass into the set \(B_{0i}\), and let
\[ B^*=\bigcup_{i=1}^{4} B_{0i}. \]
The set \(B^*\) is a \(b\)-set, and \(\operatorname{mes}B^*<\operatorname{mes}B_0\). The contradiction obtained shows that \(\beta=0\).

Thus the existence has been proved of a closed set \(B_0\subset \pi\) of measure zero, containing both bases of \(\pi\), and such that together with each of its points there belongs to it a rectilinear segment from one base

to another containing this point. At the same time, the existence of \(a\)-systems of arbitrarily small area has been proved.

Let \(\pi\) be the square \(0 \leq x \leq 1,\ 0 \leq y \leq 1\). The set \(B_0\) does not yet give the function we need, since the segments joining in it the points of the bases of the square \(\pi\) do not establish a one-to-one correspondence. But this is easy to correct.

Recall that

\[ B_0=\bigcap_{n=1}^{\infty} \widetilde A_n . \]

For the \(a\)-system \(\widetilde A_n\), denote by \(N_n^1\) and \(N_n^2\) the sets of vertices of the parallelograms entering into \(\widetilde A_n\), with \(N_n^1\) being the set of vertices situated on the lower base of \(\pi\), and \(N_n^2\) the set of vertices situated on the upper base of \(\pi\). Let

\[ N_1=\bigcup_{n=1}^{\infty} N_n^1 \quad\text{and}\quad N_2=\bigcup_{n=1}^{\infty} N_n^2 . \]

Each of the sets \(N_1\) and \(N_2\) is countable.

Remove \(N_1\) and \(N_2\), respectively, from the lower and upper bases of \(\pi\), and denote the remaining sets by \(D_1\) and \(D_2\).

For each point of \(D_1\) there exists a unique segment in \(B_0\) joining this point with some point of \(D_2\), and conversely. Denoting by \(\alpha(x)\) the angle of inclination of the segment joining the point \(x\) of \(D_1\) with \(D_2\), we note that the function \(\alpha(x)\) is continuous on \(D_1\). Put the points of the sets \(N_1\) and \(N_2\) into one-to-one correspondence in some way and join the corresponding points by segments. We obtain a one-to-one mapping of the entire lower side of the square onto the upper side. The function \(\alpha(x)\) is thereby supplemented by certain values on the countable set \(N_1\) and will thus be measurable.

Received
20 III 1962