UDC 519.53

MATHEMATICS

A. G. VITUSHKIN

PROOF OF THE UPPER SEMICONTINUITY OF THE VARIATION OF A SET

(Presented by Academician A. N. Kolmogorov on 15 VI 1965)

In this note it is proved that the (k)-dimensional variation (V_k(e)) of a closed set (e), situated in Euclidean space, is upper semicontinuous if the variations of smaller dimension are bounded. Let us recall the definition of (k)-dimensional variation (see ((^1))).

Let (E_n) be (n)-dimensional Euclidean space; (\Omega_n^k) the space of ((n-k))-dimensional planes (\beta_{n-k}) in (E_n); (\mu_n^k) the Haar measure in the space (\Omega_n^k), invariant with respect to the group of transformations of the space (\Omega_n^k) generated by motions of the space (E_n); (V_0(e)) the number of components of the set (e). The (k)-th variation of a closed set (e) is the number

[
V_k(e)=\int_{\Omega_n^k} V_0(e\cap \beta_{n-k})\,d\mu_n^k .
]

The normalization of the measure (\mu_n^k) is chosen so that the (k)-th variation of a (k)-dimensional unit cube is equal to 1.

In the space of subsets of (E_n) we shall use the deviation metric, i.e., as the distance between sets (e) and (f) from (E_n) we take the quantity
(r(e,f)=\max[\rho(e,f),\rho(f,e)]), where
(\rho(A,B)=\sup_{x\in A}\rho(x,B)), and (\rho(x,B)) is the distance from the point (x) to the set (B).

After these explanations we formulate precisely our result.

Theorem 1. If, for closed sets (e) and (e_1,e_2,\ldots,e_i,\ldots),

1) (\displaystyle \lim_{i\to\infty} r(e,e_i)=0);

2) all (V_m(e_i)) ((m=0,1,2,\ldots,(k-1);\ i=1,2,\ldots)) do not exceed some constant (M),

then

[
\sup_{i\to\infty} V_k(e_i)\geq V_k(e).
]

Consider the minimal (\sigma)-algebra (\nu_{k,n}) of sets in (E_n) containing all closed subsets of (E_n) with finite variations whose dimension is not greater than (k).

The variation (V_k(e)), extended to all sets (e\in \nu_{k,n}), turns out to be a countably additive measure of order (k), which is semicontinuous (see the formulation of the theorem) and on all polyhedra coincides with (k)-dimensional Lebesgue measure. It seems plausible that a measure satisfying the listed conditions is unique, i.e., coincides with (V_k(e)).

Notation.

[
d(g)=\sup_{a\in g,\ b\in g}\rho(a,b)
]
is the diameter of the set (g); (\partial(e)) is the boundary of the set (e); the depth of immersion of the set (e) in (g\cap\beta) will be called
[
r(e,g,\beta)=\sup \rho(x,\partial(g\cap\beta)),
]
where the least upper bound is taken over all
(x\in e\cap g\cap\beta);

[
d(e,g,\beta_p)={\operatorname{mes}_p(g\cap\beta_p)-\operatorname{mes}_p(e\cap g\cap\beta_p)}^{1/p},
]

where (\beta_p \subset E_n) is a (p)-dimensional plane;

[
m(e,g,\beta)=\min{[\Gamma(e,g,\beta)],\ [d(e,g,\beta)]},
]

[
k(g,\beta)=\Gamma(g,g,\beta)[d(g)]^{-1}.
]

Lemma 1. Let there be given a convex domain (g \subset E_n) such that (k(g,E_n)\geq \gamma>0), and a set (e\subset E_n) such that every component of the set (e\cap g) meets (\partial(g\cap E_n)).

Denote by (\Omega(\Delta_1,\Delta_2,g)) the set of hyperplanes (\beta_{n-1}) determined by the conditions: (k(g,\beta_{n-1})\geq \Delta_1) and (m(e,g,\beta_{n-1})\geq \Delta_2 m(e,g,E_n)).

There exist numbers (\varepsilon>0), (\Delta_1>0), and (\Delta_2>0), depending only on (n) and (\gamma), such that

[
\mu_n^1[\Omega(\Delta_1,\Delta_2,g)]\geq \varepsilon m(e,g,E_n).
]

Lemma 2. Let the set (e\subset E_n) be closed and bounded, and let (g_1,g_2,\ldots,g_p) be convex, pairwise disjoint sets, with

[
k(g_i,E_n)\geq \gamma>0\quad (i=1,2,\ldots,p).
]

Then for every (k=0,1,2,\ldots,n-1),

[
\mathbf{V}k(e)\geq \sum[m(e,g_i,E_n)]^k,}^{p} C_{i,k
]

where ({C_{i,k}}) are nonnegative numbers such that, for every (i),

[
\sum_{k=0}^{n-1} C_{i,k}\geq C(n,\gamma)>0
]

((C(n,\gamma)) depends only on (n) and (\gamma)).

Proof. The lemma is proved by induction on (n).

Consider the case (n=1). Among all intervals ({g_i}) intersecting one and the same component of the set (e), at most two intervals can satisfy the condition (m(e,g_i,E_1)>0), and therefore

[
\mathbf{V}0(e)\geq \sum.}^{p} [m(e,g_i,E_1)]^0\cdot \frac{1}{2
]

(Here it is assumed that if (m(e,g,E_1)=0), then ([m(e,g_i,E_1)]^0=0)); that is, for (n=1) the lemma is proved.

Now consider the general case. Recall that

[
\mathbf{V}k(e)=C(n,k)\int}\mathbf{V{k-1}(e\cap \beta)\,d\mu_n^1,
]

where (C(n,k)>0) is a normalizing constant independent of (e). By the induction hypothesis, for every plane (\beta_{n-1}) we have

[
\mathbf{V}{k-1}(e\cap \beta,})\geq \sum_i C_{i,k-1}(\beta_{n-1})[m(e,g_i,\beta_{n-1})]^{k-1
]

[
\sum_{q=1}^{n-2} C_{i,q}(\beta_{n-1})\geq C((n-1),\gamma)\quad (i=1,2,\ldots,p).
]

Here we assume that if (g_i\cap \beta_{n-1}) is empty, then (m(e,g_i,\beta_{n-1})=0).

Integrating the last inequality, we obtain

[
\frac{\mathbf{V}k(e)}{C(n,k)}\int}\mathbf{V{k-1}(e\cap \beta)\,d\mu_n^1
\geq
\sum_i \int_{\Omega_n^1} C_{i,k-1}(\beta_{n-1})[m(e,g_i,\beta_{n-1})]^{k-1}\,d\mu_n^1.
]

Put

[
C_{i,k}=C(n,k)[m(e,g_i,E_n)]^{-k}
\int_{\Omega_n^1} C_{i,k-1}(\beta_{n-1})[m(e,g_i,\beta_{n-1})]^{k-1}\,d\mu_n^1
]

[
(k=1,2,\ldots,n-1);\qquad C_{i,0}=1,\quad \text{if } g_i\cap e \text{ contains a component meeting}
]

(\partial(g_i,E_n)) components, and (C_{i,0}=0) otherwise; we have

[
V_0(e)\geq \sum_i C_{i,0}
=\sum_i C_{i,0}\,[m(e,g_i,E_n)]^0.
]

Thus, for every (k=0,1,\ldots,n-1),

[
V_k(e)\geq \sum_i C_{i,k}\,[m(e,g_i,E_n)]^{k-1},
]

and, by virtue of Lemma 1,

[
\begin{aligned}
\sum_{k=0}^{n-1} C_{i,k}
&= C_{i,0}+\sum_{k=1}^{n-1} C(n,k)
\int_{\Omega(\Delta_1,\Delta_2,g_i)}
C_{i,k-1}(\beta_{n-1})[m(e,g_i,\beta_{n-1})]^{k-1}\times\
&\quad \times [m(e,g_i,E_n)]^{-k}\,d\mu_n^1 \geq
C_{i,0}+\sum_{k=1}^{n-1} C(n,k)[m(e_i g_i,E_n)]^{-k}\times\
&\quad \times
\int_{\Omega(\Delta_1,\Delta_g,g_i)}
C_{i,k-1}(\beta_{n-1})[\Delta_2 m(e,g_i,E_n)]^{k-1}\,d\mu_n^1 \geq\
&\geq C_{i,0}+\sum_{k=1}^{n-1} C(n,k)\Delta_2^{k-1}[m(e,g_i,E_n)]^{-1}
\int_{\Omega(\Delta_1,\Delta_2,g_i)} C_{i,k}(\beta_{n-1})\,d\mu_n^1 \geq\
&\geq C_{i,0}+\min_k C(n,k)\Delta_2^{k-1}[m(e,g_i,E_n)]^{-1}
\int_{\Omega(\Delta_1,\Delta_2,g_i)}\sum_{k=1}^{n-k} C_{i,k}(\beta_{n-1})\,d\mu_n^1 \geq\
&\geq C_{i,0}+C'(n,\Delta_1)[m(e,g_i,E_n)]^{-1}\varepsilon\cdot m(e,g_i,E_n)C((n-1),\Delta_1,\gamma)\geq C(n,\gamma).
\end{aligned}
]

The lemma is proved.

Lemma 3. Let the closed set (e) be an (\varepsilon)-net for the measurable set (f\subset E_n). Then, for every (\varepsilon\leq \varepsilon(f)), for some (k=k(\varepsilon)) the inequality
[
V_k(e)\geq C(n)[\operatorname{mes}_n f-\operatorname{mes}_n e]\varepsilon^{k-n},
]
holds, where (C(n)) depends only on (n).

Proof. We shall assume that (\Delta=\operatorname{mes}_n f-\operatorname{mes}_n e>0). Fix (p) equal pairwise nonintersecting cubes (g_1,g_2,\ldots,g_p) with side (4\varepsilon) and such that

[
\sum_i \operatorname{mes}_n(f\cap g_i)\geq (1-\tfrac18\Delta)\operatorname{mes}_n f.
]

Since
[
\sum_i [m(e,g_i,E_n)]^n\geq C_1(n)\Delta,
]
then by virtue of Lemma 2 one can indicate indices (i_1,i_2,\ldots,i_q) and a number (k) such that

[
V_k(e)\geq \frac{C(n,\gamma)}{n}\sum_{s=1}^{q}[m(e,g_{i_s},E_n)]^k,
]

[
\sum_{s=1}^{q} m(e,g_{i_s},E_n)^n\geq C_2(n)\Delta,
]

and consequently,
[
V_k(e)\geq C(n)\cdot \Delta\cdot \varepsilon^{k-n}.
]

The lemma is proved.

Notation.
Let (\varphi_n^k) be the bundle of (k)-dimensional planes (\tau_k) passing through one and the same point; let (\overline m_n^k) be the measure in the space (\varphi_n^k), invariant with respect to the group of transformations generated by rotations of space; (\beta_{s+n-k}(\beta_s,\tau_k)) be the plane of dimension (s+n-k) containing the (s)-dimensional plane (\beta_s\subseteq\tau_k) and the ((n-k))-dimensional plane (\beta_{n-k}\subset E_n), which is ...

orthogonal complement to (\tau_k)

[
V_{s,\tau_k}(e)=\int_{\beta_s\subset\tau_k} V_0\bigl(e\cap \beta_{s+n-k}(\beta_s,\tau_k)\bigr)\,d\mu_k^{\,k-s}.
]

Lemma 4. Let the set (f\subset E_n) be closed and such that, for every plane (\beta_s\subset\tau_k) ((\tau_k) fixed) and for every pair of components (f_1) and (f_2) of the set (f\cap \beta_{s+n-k}(\beta_s,\tau_k)),

[
\min_{\substack{a\in f_1\ b\in f_2}}\rho(a,b)\ge \Delta>0
]

((\Delta) does not depend on (\beta_s,f_1,f_2)), and let the set (e) be an (\varepsilon)-net for (f) and lie in the ((\tfrac14\Delta))-neighborhood of the set (f). Then, if (\varepsilon<\varepsilon(f)) ((\varepsilon(f)) does not depend on (e)), then

[
\sum_{s=0}^{k-1} V_{s,\tau_s}(e)\ge C(k)\,[V_{k,\tau_k}(f)-V_{k,\tau_k}(e)]\,\varepsilon^{-1}
]

((C(k))—see Lemma 3).

Lemma 5. If a closed set is equal to (\lim_{i\to\infty} e_i), then for every plane (\tau_k)

[
\lim_{i\to\infty}\sum_{s=0}^{k-1} V_{s,\tau_k}(e_i)=+\infty,
]

provided only that

[
\inf_{i\to\infty} V_{k,\tau_k}(e_i)