Mathematics

A. G. VITUSHKIN, L. D. IVANOV, M. S. MELNIKOV

INCOMMENSURABILITY OF THE MINIMAL LINEAR MEASURE WITH THE LENGTH OF A SET

(Presented by Academician A. N. Kolmogorov on 14 II 1963)

In the work of A. N. Kolmogorov \((^{1})\) the question is posed of the commensurability of the Hausdorff length of a set \(l(M)\) with the minimal linear measure of the set \(L(M)\), i.e., whether the inequality

\[ l(M)\leq C L(M) \]

holds for every set \(M\) in the plane of finite positive length, where \(C\) is a positive absolute constant. In this note an example is constructed of a set \(M\) in the plane of finite positive length and with \(L(M)=0\).*

Notation and construction of the set. \(d(M)\) is the diameter of the set \(M\); \(c(M)\) is the convex hull of the set \(M\);

\[ c(M)=\inf \sum_{\mu} d(S_\mu) \]

over all coverings \(\{S_\mu\}\) of the set \(M\) by convex domains \(S_\mu\).

Further, let \(k\) be a positive integer; \(\delta\) a positive number; \(\tau\) a number equal to 0 or 1. \(P_\tau(k,\delta,R)\) is a transformation which sends the horizontal segment \(R=\{a\leq x\leq b,\ y=c\}\) into the set consisting of \(2k\) horizontal segments of the form

\[ \left\{a+\frac{(i-1)(b-a)}{k}\leq x\leq a+\frac{(2i-1)(b-a)}{2k},\quad y=c+\delta\tau\right\}, \]

where \(i=1,2,\ldots,k\).

If the set \(M\) consists of \(m\) horizontal segments \(r_l\), then

\[ P_\tau(k,\delta,M)=\bigcup_{l=1}^{m} P_\tau(k,\delta,r_l). \]

Let

\[ M_0=\{0\leq x\leq 1,\ y=0\}, \]

\[ M(\tau_1)=P_{\tau_1}(k_0,\delta_0,M_0), \]

\[ \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots \]

\[ M(\tau_1,\ldots,\tau_j) = P_{\tau_j}\bigl(k(\tau_1,\ldots,\tau_{j-1}),\delta(\tau_1,\ldots,\tau_{j-1}),M(\tau_1,\ldots,\tau_{j-1})\bigr). \]

Thus the set \(M(\tau_1,\ldots,\tau_j)\) consists of

\[ k_0\cdot k(\tau_1)\cdots k(\tau_1,\ldots,\tau_{j-1}) \]

horizontal segments \(r_l(\tau_1,\ldots,\tau_j)\), lying on one straight line; all of them have the same length

\[ l\bigl(r_l(\tau_1,\ldots,\tau_j)\bigr) = \frac{1}{2^j k_0\cdots k(\tau_1,\ldots,\tau_{j-1})}. \]

Denote

\[ M(\tau_1,\ldots,\tau_j)^i = \bigcup_{\tau_{j+1},\ldots,\tau_{j+i}=0;\,1} M(\tau_1,\ldots,\tau_j,\tau_{j+1},\ldots,\tau_{j+i}). \]

Let \(J_i(\tau_1,\ldots,\tau_j)\) be the number of segments in the set \(r_l(\tau_1,\ldots,\tau_j)^i\)

\[ J_i(\tau_1,\ldots,\tau_j) = 2^i \sum_{\tau_{j+1},\ldots,\tau_{j+i-1}=0;\,1} k(\tau_1,\ldots,\tau_j)\cdots k(\tau_1,\ldots,\tau_{j+i-1}), \]

Now we construct the set \(M^{N_1}\), having first imposed on the numbers

\[ k_0,\ldots,k(\tau_1,\ldots,\tau_{N_1-1}),\quad \delta_0,\ldots,\delta(\tau_1,\ldots,\tau_{N_1-1}), \]

* A. N. Kolmogorov expressed the supposition of the equality of these measures: “Es scheint mir nicht unwahrscheinlich, daß immer \(m_1(E)=\mu_1(E)\)” \((^{1})\), p. 361).

where \(\tau_1,\ldots,\tau_{N_1-1},\ \tau_{N_1}=0;\ 1\), the following conditions:

a) \(N_1J_{N_1-1}(0)\leq k(1)\),

\[ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot \]

\[ N_1J_{N_1-j}(\tau_1,\ldots,\tau_{j-1},0)\leq k(\tau_1,\ldots,\tau_{j-1},1). \]

\[ j=1,\ldots,N_1-1; \]

b) \(N_1\delta_0\leq l(r_l(\tau_1))\),

\[ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot \]

\[ N_1\delta(\tau_1,\ldots,\tau_{j-1})\leq l(r_l(\tau_1,\ldots,\tau_j)). \]

\[ j=1,\ldots,N_1; \]

c) \(\delta_0\geq 2^3l(r_l(\tau_1,\tau_2))\),

\[ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot \]

\[ \delta(\tau_1,\ldots,\tau_{j-1})\geq 2^{j+2}l(r_l(\tau_1,\ldots,\tau_j,\tau_{j+1})) \]

\[ j=1,\ldots,N_1-1. \]

Thus, we have constructed the set \(M^{N_1}\), consisting of a finite number \(J_{N_1}\) of intervals \(r_l(\tau_1,\ldots,\tau_{N_1})\). Now we construct the set \(M^{N_1N_2}\), applying to each interval in \(M^{N_1}\) the operation \(P_\tau\) \(N_2\) times with the same conditions a), b), only for the number \(N_2\), while in condition c) we increase the exponents of the power of \(2\) by \(N_1\): \(\delta_0^1\geq 2^{3+N_1}l(r_l(\tau_1,\tau_2))\). By \(\delta^\nu(\tau_1,\ldots,\tau_j)\), \(k^\nu(\tau_1,\ldots,\tau_j)\) are denoted the numbers \(\delta,k\) for the \((\nu+1)\)-st step of the construction. Further, applying to each interval of the set \(M^{N_1N_2}\) the operation \(P_\tau\) \(N_3\) times, with in condition c) the numbers \(\delta^2(\tau_1,\ldots,\tau_j)\) involving exponents of the form \(3+N_1+N_2,\ 4+N_1+N_2,\ldots,\ N_3+N_2+N_1+1\), we construct the set \(M^{N_1N_2N_3}\), and so on. We obtain the set \(M^{N_1\cdots N_k}\). Thus, let a sequence of natural numbers \(N_1,\ldots,N_k,\ldots\) be given, the growth conditions on which will be imposed below; then

\[ M=\lim_{k\to\infty} M^{N_1\cdots N_k}. \]

Lemma 1. Let \(\bar d(E)=\sup\max\{|x_2-x_1|,\ |y_2-y_1|\}\) over all \(\{x_1,y_1\},\ \{x_2,y_2\}\in E;\ S\) be a set. Then

\[ \left(2+\frac{1}{2^j}+\cdots+\frac{1}{2^{j+i+1}}\right)\bar d(S)\geq l\bigl(S\cap M(\tau_1,\ldots,\tau_j)^i\bigr); \]

\[ j=0,1,\ldots;\quad i=0,1,\ldots . \]

Lemma 2. The coverings of the set \(M^{N_1}\) and of the sets \(M^{N_1\cdots N_n}\), where \(n=1,2,\ldots\),

\[ c(M^{N_1})\geq \frac{1}{4\sqrt2}; \qquad c(M^{N_1\cdots N_n})\geq \frac{1}{4\sqrt2}. \]

This lemma follows at once from Lemma 1, taking into account that \(d(S)\geq \frac1{\sqrt2}\bar d(S)\), since condition c) for \(M^{N_1\cdots N_n}\) coincides with condition c) for \(M^{N_1+\cdots+N_n}\).

Lemma 3. The length of the set \(M\)

\[ l(M)\geq \frac{1}{4\sqrt2}. \]

The lemma follows immediately from Lemma 2:

\[ l(M)\geq c\left(\lim_{k\to\infty}M^{N_1,\ldots,N_k}\right)\geq \lim_{k\to\infty}c\left(M^{N_1\cdots N_k}\right)\geq \frac{1}{4\sqrt2}. \]

Lemma 4. For any contraction mapping \(f\) of the set \(M^i\) onto a straight line, the measure of the image

\[ \operatorname{mes}(f(M^i))\leq \frac{20}{i}\,l(M^i),\quad \text{where } i\leq N_1. \]

We prove this by induction on \(i\). For \(i\leq 20\) the assertion of the lemma is obvious. Suppose the lemma is true for all \(M^i,\ i=1,2,\ldots,n\leq N_1\).

We note that

\[ M^{n+1}=M(0)^n \bigcup M(1)^n,\qquad f(M^{n+1})=f(M(0)^n)\bigcup f(M(1)^n), \]

then

\[ \operatorname{mes}(f(M^{n+1}))=\operatorname{mes}(f(M(0)^n))+\operatorname{mes}(f(M(1)^n))- \]

\[ -\operatorname{mes}\bigl(f(M(0)^n)\cap f(M(1)^n)\bigr). \tag{1} \]

1. If

\[ \operatorname{mes}(f(M(0)^n))\leq \frac{9}{10}\frac{20}{n}\,l(M(0)^n), \]

then, since

\[ l(M(0)^n)=l(M(1)^n)=\frac12 l(M^{n+1})=\frac12, \]

we obtain

\[ \operatorname{mes}(f(M^{n+1}))\leq \frac{9}{10}\frac{20}{n}\frac12+\frac{20}{n}\frac12 \leq \frac{20}{n+1}\,l(M^{n+1}), \]

since \(n>20\).

1. Let

\[ \operatorname{mes}(f(M(0)^n))>\frac{9}{10}\frac{20}{n}\,l(M(0)^n)=\frac{9}{n}. \]

The set \(f(M(0)^n)\) on the line consists of no more than \(k_0J_n(0)\) intervals or points, since \(f(M(0)^n)\) is a contracted image of the set \(M(0)^n\), which consists of the same number of intervals.

Let

\[ A_p=\inf(f(r_p(0)^n)),\qquad A'_p=\inf(f(r_p(1)^n)), \]

\[ B_p=\sup(f(r_p(0)^n)),\qquad B'_p=\sup(f(r_p(1)^n)); \]

then, since the distance from \(r_p(0)^n\) to \(r_p(1)^n\) is equal to \(\delta_0\),

\[ |A'_p-A_p|\leq \delta_0,\qquad |B'_p-B_p|\leq \delta_0, \]

i.e.

\[ \operatorname{mes}\bigl(f(r_p(0)^n)\cap A'_pB'_p\bigr)\geq \operatorname{mes}(f(r_p(0)^n))-2\delta_0. \tag{2} \]

Further,

\[ f(M(1)^n)=\bigcup_{p=1}^{k_0} f(r_p(1)^n) = \bigcup_{p=1}^{k_0}\bigcup_{l=1}^{k(1)} f\bigl(r_{pl}(1,0)^{\,n-1}\cup r_{pl}(1,1)^{\,n-1}\bigr). \]

Since the distance between two neighboring subsets of the form
\(f(r_{pl}(1,0)^{\,n-1}\cup r_{pl}(1,1)^{\,n-1})\), of which the set \(f(r_p(1)^n)\) consists, is not greater than
\(l(r_{pl}(1,0))=l(M(1)^n)/2k_0k(1)\), it follows that into any interval \(CD:\ A'_p\leq C\leq D\leq B'_p\) there fall no fewer than
\(|D-C|/2l(r_{pl}(1,0))-1\) of these subsets. It is easy to see that into two intersecting intervals \(CD, EF:\ C<E<D<F\),
\(A'_{p_1}\leq C<D\leq B'_{p_1}\) and \(A'_{p_2}\leq E<F\leq B'_{p_2}\), there fall no fewer than
\(|C-F|/2l(r_{pl}(1,0))-2\) of these subsets. Hence, taking account of (2) and of the fact that the set \(f(r_p(0)^n)\) consists of no more than \(J_n(0)\) intervals, we obtain that this set \(f(r_p(0)^n)\) “covers” no fewer than

\[ \frac{\operatorname{mes}(f(r_p(0)^n))-2\delta_0}{2l(r_{pl}(1,0))} \]

\[ -\,J_n(0) \]

subsets of the form
\(f(r_{pl}(1,0)^{\,n-1}\cup r_{pl}(1,1)^{\,n-1})\).
This means that all of \(f(M(0)^n)\) “covers” no fewer than

\[ \frac{\operatorname{mes}(f(M(0)^n))-2\delta_0k_0}{2l(r_{pl}(1,0))} -k_0J_n(0) \geq \frac{\frac{9}{10}\frac{20}{n}\,l(M(0)^n)-2\delta_0k_0}{2l(r_{pl}(1,0))} -k_0J_n(0) \tag{3} \]

of these subsets. Taking (1), (3) into account, and

\[ \operatorname{mes}\left(f\left(r_{pl}(1,0)^{\,n-1}\cup r_{pl}(1,1)^{\,n-1}\right)\right) \leq \frac{2\cdot 20}{n-1}\, l\left(r_{pl}(1,0)\right), \]

we have

\[ \begin{aligned} \operatorname{mes}\left(f\left(M^{n+1}\right)\right) &\leq \frac{20}{n}\,\frac12+\frac{20}{n-1}\cdot 2l\left(r_{pl}(1,0)\right)\times\\ &\quad \times\left[ k_0k(1)-\frac{\frac{9}{n}\,l\left(M(0)^n\right)-k_0\delta_0}{l\left(r_{pl}(1,0)\right)} +k_0J_n(0) \right] \\ &= \frac{20}{n}\,\frac12+\frac{20}{n-1}\,\frac12 -\frac{20\cdot 9}{(n-1)n} +\frac{40}{n-1}\,\delta_0 k_0 +\frac{40}{n-1}\,l\left(r_{pl}(1,0)\right)k_0J_n(0). \end{aligned} \]

From a) it follows that \(l(r_{pl}(1,0))k_0J_n(0)\leq 1/N_1\); from b) it follows that \(2\delta_0k_0\leq 1/N_1\), whence

\[ \operatorname{mes}\left(f\left(M^{n+1}\right)\right) \leq \frac{10}{n}+\frac{10}{n-1}+\frac{20}{(n-1)N_1} +\frac{40}{(n-1)N_1} -\frac{180}{n(n-1)} \leq \frac{20}{n+1}, \]

where \(20<n\leq N_1\), as was required.

We now return to the conditions on the growth of the sequence \(N_1,N_2,\ldots,N_k,\ldots\). Let a sequence of positive \(\varepsilon_1,\varepsilon_2,\ldots,\varepsilon_k,\ldots\), tending to zero, be given. It follows from Lemma 4 that if only \(N_1>20\cdot 4/\varepsilon_1\), then for any contracted mapping \(f\) onto a line
\(\operatorname{mes}(f(M^{N_1}))\leq \varepsilon_1/4\). Denote by \(\overline E_{\sigma_1}\) the \(\sigma_1\)-extension of the set \(E\). Since the set \(M^{N_1}\) consists of a finite number of intervals \(J_{N_1}\), for \(\sigma_1>0\) and \(4J_{N_1}\sigma_1<\varepsilon_1\) the measure of the \(\sigma_1\)-extension of the image of the set \(M^{N_1}\)

\[ \operatorname{mes}\left(\overline{f\left(M^{N_1}\right)}_{\sigma_1}\right)<\varepsilon_1. \]

Next, choose

\[ N_2>\max\left\{\frac{J_{N_1}\cdot 80}{\varepsilon_2},\,\frac{1}{2\sigma_1}\right\}; \]

it is immediately clear that \(\operatorname{mes}(f(M^{N_1N_2}))<\varepsilon_2/4\) for any contracted mapping \(f\).

Find \(\sigma_2<\frac14\sigma_1\) and

\[ \operatorname{mes}\left(\overline{f\left(M^{N_1N_2}\right)}_{\sigma_2}\right)\leq \varepsilon_2, \]

and so on.

In general, the condition on the growth of \(N_1,\ldots,N_k,\ldots\) is

\[ \text{g)}\quad N_k>\max\left\{ \frac{J_{N_1\ldots N_{k-1}}\cdot 80}{\varepsilon_k},\, \frac{1}{2\cdot\sigma_{k-1}} \right\}. \]

Lemma 5. Let

\[ M=\lim_{k\to\infty} M^{N_1\ldots N_k}, \]

where the sequence \(N_1,\ldots,N_k,\ldots\) satisfies condition g). Then \(\operatorname{mes}(f(M))=0\) for any contracted mapping \(f\) onto a line.

Since \(N_k\leq \dfrac{1}{2\sigma_{k-1}}\), it follows, by condition b), that

\[ \delta_0^k\leq \frac{1}{2N_k}\leq \frac14\sigma_{k-1}, \qquad \delta_0^{k+1}\leq \frac14\sigma_k\leq \frac1{16}\sigma_{k-1}; \]

then the maximal deviation of the set \(M^{N_1\ldots N_k}\) from \(M^{N_1\ldots N_{k-1}}\) is less than \(\frac12\sigma_{k-1}\), and that of the set \(M^{N_1\ldots N_{k+1}}\) from \(M^{N_1\ldots N_{k-1}}\) is less than \((\frac12+\frac14)\sigma_{k-1}\), and, in general,

\[ M\subset \left(M^{N_1\ldots N_{k-1}}\right)_{\sigma_{k-1}}. \]

But

\[ \operatorname{mes}\left(\overline{f\left(M^{N_1\ldots N_{k-1}}\right)}_{\sigma_{k-1}}\right)\leq \varepsilon_{k-1}. \]

Hence, by Whitney’s theorem on the extension of a function with Lipschitz condition with constant 1 on the set \(M\) to the whole plane with Lipschitz constant \(C\), it follows that for any contracted mapping onto a line

\[ \operatorname{mes}(f(M))\leq C\varepsilon_{k-1},\qquad \text{i.e.}\quad \operatorname{mes}(f(M))=0. \]

It follows from Lemmas 3 and 5 that for the set \(M\) the length \(l(M)\) and the minimal linear measure \(L(M)\) are incommensurable.

Received
24 X 1962

CITED LITERATURE

1. A. N. Kolmogoroff, Math. Ann., 107, 351 (1932).