Exercise 21, § 8

Denote by $\lambda$ the Lebesgue measure on $\mathbf{R}^n$.

a) Let $C_0$ be a cube (GT, VI, §1, No. 1) in $\mathbf{R}^n$, and let u be a numerical function on $C_0$ that is integrable (for $\lambda$). Let s be a real number such that

Show that there exists a sequence $(I_k)$ of open cubes contained in $C_0$, pairwise disjoint, and satisfying the following conditions:

$1^\circ$ $|u(x)| \leq s$ almost everywhere in the set $C_0 - \bigcup_k I_k$;

$2^\circ$ if one sets

then $u_k \leq 2^n s$ for all $k$;

$3^\circ$ $\sum_k \lambda(I_k) \leq \frac{1}{s} \int_{C_0} |u(x)| \, d\lambda(x)$.

(Dividing into two equal intervals each of the projections of $C_0$, among the $2^n$ pairwise disjoint open cubes choose the cubes $I_{1i}$ such that the corresponding 'mean values' $u_{1i}$ are all $\geq s$, and, because of the choice of s, show that $u_{1i} \leq 2^n s$. Similarly decompose into $2^n$ cubes each of the cubes of the first subdivision that does not figure among the $I_{1i}$, which yields a second finite family of cubes $I_{2i}$ chosen among all of the new cubes obtained as those for which the corresponding mean value $u_{2i}$ is $\leq s$. Continue by recursion. To prove that the condition $1^\circ$ is verified, argue by contradiction, noting that the subset $B_m$ of $C_0 - \bigcup_k I_k$ where $|u(x)| \geq s + 1/m$ is, up to a negligible set, the intersection of a decreasing sequence $(U_r)$ of open sets, where $U_r$ is the union of the cubes of the r-th subdivision that do not figure among the $I_k$ and that contain a point of $B_m$.)

b) Let $\mathscr{F}$ be the set of integrable numerical functions u on $C_0$ having the following property: for every cube $C \subset C_0$, if $m_C(u)$ is the mean value

then $\int_C |u(x) - m_C(u)| \, d\lambda(x) \leq \lambda(C)$; if $u \in \mathscr{F}$, the same is true of $u - c$ for every real number c. For every function $u \in \mathscr{F}$, every number $\sigma > 0$ and every cube $C \subset C_0$, denote by $S(\sigma, u, C)$ the set of $x \in C$ such that $|u(x) - m_C(u)| \geq \sigma$. Finally, denote by $G(\sigma)$ the smallest number such that

as $u$ runs over $\mathscr{F}$, and C over the set of cubes contained in $C_0$; one has $G(\sigma) \leq 1/\sigma$.

Show that if $\sigma/2^n > s \geq 1$ then

(One can restrict oneself to proving that, for a $u \in \mathscr{F}$, one has

and one can suppose, to simplify, that $m_{C_0}(u) = 0$. Make use then of the decomposition of $C_0$ into a union of cubes $I_k$ and a set where $|u(x)| \leq s$ almost everywhere, on noting that if $x \in I_k$ belongs to $S(\sigma, u, C_0)$, then $|u(x) - m_{I_k}(u)| \geq \sigma - 2^n s$ and consequently

c) Deduce from b) that there exist two constants $B, b$ depending only on $n$, such that, for every function $u \in \mathscr{F}$ and every number $\sigma > 0$, one has $\lambda(S(\sigma, u, C_0)) \leq Be^{-b\sigma} \lambda(C_0)$. (Take $s = e$ in b).)