MATHEMATICS

E. V. GLIVENKO

ON A MEASURE OF HAUSDORFF TYPE

(Presented by Academician P. S. Aleksandrov, 5 IX 1956)

A considerable class of measures for sets situated in Euclidean space or in a metric space with a countable base is obtained by the following construction. Choose a countable system of open sets (\Gamma_i) such that their diameter (\alpha(\Gamma_i) \to 0) as (i \to \infty). Let the regions (\Gamma_i) cover the space in the sense of Vitali.* Assign to each of the regions (\Gamma_i) an arbitrary number (\alpha_i \geq 0), which we shall call the elementary measure (\operatorname{mes}\mathrm{e}\Gamma_i) of the region (\Gamma_i). Fix an arbitrary (\varepsilon > 0). Cover the set (M) by regions (\Gamma, \ldots) from the system ({\Gamma_i}) such that}, \Gamma_{i_2

[
d(\Gamma_{i_j}) < \varepsilon \quad (j = 1, 2, \ldots).
]

Consider

[
\sum_{j=1}^{\infty} \alpha_{i_j}.
]

The greatest lower bound of such sums over all possible coverings of the set (M) by regions from ({\Gamma_i}) will be called the (\varepsilon)-measure (\operatorname{mes}_{\varepsilon} M) of the set (M). The limit of these measures as (\varepsilon) tends to zero will be called the measure of the set (M) of Hausdorff type and denoted by (\operatorname{mes} M). A special case of such a measure is the ordinary linear Hausdorff measure in the plane, when the system ({\Gamma_i}) consists of all disks with rational coordinates of their centers and rational radii, and the elementary measure of each disk is its diameter.

In order that the measure be convenient in use, it is necessary that it possess the so-called (F)-property, which consists in the following:

a) if the set (M) has finite measure, then for any (\varepsilon > 0) there exists a closed set (F \subset M) such that (|\operatorname{mes} M - \operatorname{mes} F| < \varepsilon);

b) if the set (M) has infinite measure, then for any (N > 0) there exists a closed set (F \subset M) such that (\operatorname{mes} F > N).

The (F)-property for the ordinary Hausdorff measure as applied to the sets (F_{\sigma\delta}) was proved in 1952 by A. S. Besicovitch ((^1)). Earlier, R. A. Minlos ((^2)) proved the (F)-property for (A)-sets in the case of one special measure, but, in essence, his proof is applicable to the ordinary Hausdorff measure. In the case where the whole space is representable as the sum of a countable number of sets of finite measure, the (F)-property was also proved for measures of Hausdorff type ((^3)).

The present work is devoted to the proof of the (F)-property for measures of Hausdorff type in the general case as applied to (A)-sets.

Theorem 1. Let, in a metric space with a countable base, some measure of Hausdorff type be introduced. Then:

a) If some (A)-set (M), situated in the space (R), has finite measure, then, whatever (\varepsilon > 0) may be, there exists a closed set (F \subset M) such that (|\operatorname{mes} F - \operatorname{mes} M| < \varepsilon).

b) If some (A)-set (M) has infinite measure, then, whatever the number (N) may be, there exists a closed set (F \subset M) such that (\operatorname{mes} F > N).

* This means that for any point (\xi) of the space and for any number (\varepsilon > 0) there exists an open set from the system ({\Gamma_i}), covering the point (\xi), whose diameter is less than (\varepsilon).

Part a) does not differ from the case considered in (3), when the whole space has finite measure. Thus, part b) remains to be proved. This proof is based on the following lemma:

Lemma. If a set (M \subset R) is the sum of an increasing sequence of measurable sets
[
M_1 \subset M_2 \subset \ldots \subset M_n \subset \ldots,\qquad
M=\sum_{i=1}^{\infty} M_i,
]
then
[
\lim_{n\to\infty}\operatorname{mes}{\varepsilon} M_n \geq \operatorname{mes} M.
]

Proof. Two cases are possible:
[
1)\ \lim_{n\to\infty}\operatorname{mes}{\varepsilon} M_n=\infty,\qquad
2)\ \lim M_n<\infty.}\operatorname{mes}_{\varepsilon
]

In the first case the assertion of the lemma is obvious. Let us consider the second case. Fix (\delta>0). Cover each of the sets (M_i) by a system (g_i) of domains so that the sum of the elementary measures of these domains differs from the measure (\operatorname{mes}{\varepsilon} M_i) by less than (\delta/2^i). From each covering (g_i) choose a finite system (g_i^) so that the sum of the elementary measures of the domains from the system (g_i^) differs from the sum of the elementary measures of the domains from the system (g_i) by less than (\delta/2^i). Each covering (g_i^) consists of a finite number (n_i) of domains
[
G_1^{i},\ldots,G.}^{i*
]

Consider the covering (g_1^). There is a subsequence of the natural sequence (\alpha_{11}=1,\alpha_{12},\ldots,\alpha_{1n},\ldots) such that each domain of the system (g_1^) either belongs to all the coverings (g_{\alpha_{1i}}^), or does not belong to any of them for (i>1). Consider the system (g_{\alpha_{12}}^). For it there is likewise a sequence of numbers (\alpha_{21}=\alpha_{12},\alpha_{22},\alpha_{23},\ldots), which is a subsequence of the first sequence ({\alpha_{1i}}), and each domain of the system (g_{\alpha_{12}}^) either belongs to each of the systems (g_{\alpha_{2i}}^) ((i>1)), or does not belong to any one of them. Analogous subsequences are found for each system (g_{\alpha_{i1}}). As a result we obtain a sequence of sequences, each subsequent one of which is a subsequence of the preceding one.

Consider the diagonal sequence ({\alpha_{i1}}), and henceforth we shall deal only with the sets and coverings from this sequence. Let ({\Gamma_{i_k}^0}) be the system of domains from ({\Gamma_i}), each of which belongs to all coverings (g_{\alpha_{i1}}^), beginning with some (i). Now remove from each covering (g_{\alpha_{i1}}^) the domains that have entered the system ({\Gamma_{i_k}^0}). Denote the remaining system of domains by (\widetilde g_{\alpha_{i1}}).

We shall now consider only the sets
[
\widetilde M_{\alpha_{i1}}=M_{\alpha_{i1}}\cap g_{\alpha_{i1}}
]
and their coverings (\widetilde g_{\alpha_{i1}}). Consider the set (\widetilde M_{\alpha_{11}}) and its covering (\widetilde g_{\alpha_{11}}). Consider also the covering (\widetilde g_{\alpha_{1+q,\,1}}) for some (q). The domains of the covering (\widetilde g_{\alpha_{1+q,\,1}}), with respect to the domains of the covering (\widetilde g_{\alpha_{11}}), are divided into three groups. In the first group (A_{\alpha_{1+q,\,1}}^1) we include those domains of the covering (\widetilde g_{\alpha_{1+q,\,1}}) which are contained entirely inside the covering (\widetilde g_{\alpha_{11}}). In the second group (B_{\alpha_{1+q,\,1}}^1) we include the domains from the covering (\widetilde g_{\alpha_{1+q,\,1}}) which contain both points of (\widetilde g_{\alpha_{11}}) and points of the complement of (\widetilde g_{\alpha_{11}}). And in the third group (C_{\alpha_{1+q,\,1}}^1) we include the domains remaining in the system (\widetilde g_{\alpha_{1+q,\,1}}).

Consider the system of domains (A_{\alpha_{1+q,\,1}}^1). It can be shown that there is such a sufficiently large number (q_0) that the sum of the elementary measures of the domains (A_{\alpha_{1+q,\,1}}^1)

differs from the sum of the elementary measures of the system (\tilde g_{\alpha_{11}}) by less than
((2\delta/2^{\alpha_{11}}+2\delta/2^{\alpha_{1+q,1}})).

Let us now consider the system of domains

[
\tilde g_{\alpha_{i_1},1}=\tilde g_{\alpha_{11}}+B^1_{\alpha_{1+q_0},1}+C^1_{\alpha_{1+q_0},1}.
]

We carry out with the system of domains (\tilde g_{\alpha_{i_1},1}) the same operation that we have just carried out with the system (\tilde g_{\alpha_{11}}=\tilde g_{\alpha_{11}}). In this way we obtain a system of domains (\tilde g_{\alpha_{i_2},1}), and so on. Suppose we have a domain (\tilde g_{\alpha_{i_n},1}). Construct the system of domains (\tilde g_{\alpha_{i_{n+1}},1}). Consider the set (\widetilde M_{\alpha_{i_n},1}) and its covering (\tilde g_{\alpha_{i_n},1}). Consider also the covering (\tilde g_{\alpha_{i_n+q},1}) for some (q). The domains of the covering (\tilde g_{\alpha_{i_n+q},1}), in relation to the domains of the covering (\tilde g_{\alpha_{i_n},1}), are divided into three groups. In the first group (A^{i_n}{\alpha},1}) we include those domains of the covering (\tilde g_{\alpha_{i_n+q},1}) which are contained entirely inside the covering (\tilde g_{\alpha_{i_n},1}). In the second group (B^{i_n{\alpha},1}) we include those domains from the covering (\tilde g_{\alpha_{i_n+q},1}) which contain both points of (\tilde g_{\alpha_{i_n},1}) and points of the complement of (\tilde g_{\alpha_{i_n},1}). In the third group (C^{i_n{\alpha).},1}) we include the remaining domains in the system (\tilde g_{\alpha_{i_n+q},1

Consider the system of domains (A^{i_n}{\alpha},1}). There is such a sufficiently large number (q) that the sum of the elementary measures of the domains (A^{i_n{\alpha) by less than},1}) differs from the sum of the elementary measures of the system (\tilde g_{\alpha_{i_n},1
((2\delta/2^{\alpha_{i_n},1}+2\delta/2^{\alpha_{i_n+q},1})).

Denote by

[
\tilde g_{\alpha_{i_{n+1}},1}=\tilde g_{\alpha_{i_n},1}+B^{i_n}{\alpha.},1
]

We obtain a sequence of systems of domains

[
\left{\tilde g_{\alpha_{i_1},1}\subset \tilde g_{\alpha_{i_2},1}\subset \cdots \subset \tilde g_{\alpha_{i_n},1}\subset \cdots \right}.
]

Let us now consider

[
\sum_{j=1}^{\infty}\tilde g_{\alpha_{i_j},1}=T.
]

The sum of the elementary measures of the domains (T) differs from the limit of the sums of the elementary measures of the domains (\tilde g_{\alpha_i,1}) by no more than (8\delta).

At the beginning of the proof we discarded from the sets (M_i) those parts which did not enter the finite coverings (g_i^*). Each time we thereby discarded parts of the sets (M_i), the (\varepsilon)-measure of which did not exceed (\delta/2^i).

Consider now

[
\sum_{i=1}^{\infty} M_i \setminus \widetilde M_i.
]

This set will have (\varepsilon)-measure not exceeding (2\delta). Cover

[
\sum_{i=1}^{\infty} M_i \setminus \widetilde M_i
]

by a system of domains (P), the sum of whose elementary measures does not exceed (2\delta).

Let now ({\Gamma_i'}=S+T+P). The system ({\Gamma_i'}) will cover the set (M). The sum of the elementary measures of the regions of this system does not exceed (12\delta), and, since (\delta) is arbitrary, the lemma is proved.

Proof of the theorem. Map our metric space (R) by a homeomorphic transformation into the Hilbert space (K). Under this transformation our set (M), which is an (A)-set in the metric space, passes into some (A)-set (M^1) in the Hilbert space (K) (4). Define a measure of Hausdorff type in the Hilbert space by assigning to the images of the regions (\Gamma_i) the elementary measure of the (\Gamma_i) themselves. Then the measure of the set (M^1) in the Hilbert space will be equal to the measure of the set (M) in the space (R).

We shall prove the theorem for the set (M^1) in the space (K). Thereby the theorem will be proved for the sets (M) in the space (R). Construct the topological product of the space (K) with a one-dimensional Euclidean space. Denote this product by (L). Let our set (M^1) be obtained as the projection into the space (K) of some (G_\delta) situated in the space (L). Let our (G_\delta), in turn, be the intersection of open sets (G_i). Each open set (G_i) we can represent in the form of a sum of a countable number of increasing closed sets (f_{ij}). Denote the projection of the set (f_{ij}) in the space (R) by (F_{ij}); (\prod_i F_{ij}) will be a closed set contained in the set (M).

Now consider the intersection (F_{ij}\cap M^1). Fix the numbers (\varepsilon_1) and (\varepsilon_2). According to the lemma, in each (G_i) we can choose such a closed set (f_{ij_0}) that

[
\left|\operatorname{mes}{\varepsilon_1} F.}\cap M-\operatorname{mes}_{\varepsilon_1} M\right|<\frac{\varepsilon_2}{2^i
]

Since (\prod_i (F_{ij_0}\cap M)=\prod_i F_{ij_0}), it follows that (\prod_i (F_{ij_0}\cap M)) is a closed set. Since for a closed set one can choose a finite subcover from any cover, it is true that if a closed set is the intersection of a sequence of sets, then the (\varepsilon)-measure of the intersection is equal to the limit of the (\varepsilon)-measures of the intersecting sets.

Fix (N). Choose (\varepsilon_1) so that (\operatorname{mes}_{\varepsilon_1} M>2N), and (\varepsilon_2<N/2). Then

[
\operatorname{mes}{\varepsilon_1}\prod_i F>
\operatorname{mes}{\varepsilon_1}\prod_i F>N.
]

Thus the theorem is proved.

It still does not follow from this theorem that in every (A)-set of infinite measure one can choose a closed set of arbitrarily large finite measure. For the ordinary Hausdorff measure in (n)-dimensional Euclidean space this can be proved. Namely, the following theorem holds:

Theorem 2. Let (M) be an (A)-set of infinite Hausdorff measure (\mu). Then for every (N) there exists a closed set (F\subset M) such that (\mu F=N).

Received
21 VI 1956

References

(^1) A. S. Besicovitch, Proc. Kon. Ned. Akad. v. Wet. Amsterdam, ser. A, 55, No. 3, 339 (1952).
(^2) R. L. Minlos, DAN, 81, No. 5 (1951).
(^3) M. E. Munroe, Introduction to Measure and Integration, Cambridge, 1953.
(^4) F. Hausdorff, Set Theory, 1937, p. 220.