For $W=\cup_{U\in\mathcal U} U$ show that there exists $U_1,\dots,U_n: \ \sum_{i=1}^n \lambda(U_i) > \frac{1 - \epsilon}{3^d}\lambda(W)$ I'm trying to solve the following, which is Exercise 13.15 in the book Probability Theory by A. Klenke.
Let $C \subset \mathbb R^d $ be an open, bounded and convex set and assume that
$$
\mathcal{U} \subset \left \{ x+rC: x \in \mathbb R^d, r>0 \right \}
$$
is such that
$$
W:= \bigcup_{U \in \mathcal{U}}U 
$$
has finite Lebesgue measure $\lambda(W)$. Show that for any $\epsilon > 0$ there exists finitely many pairwise disjoint sets $U_1, \dots, U_n$ such that
$$
\sum_{i=1}^n \lambda(U_i) > \frac{1 - \epsilon}{3^d}\lambda(W). \tag 1
$$
Show by a counterexample that the condition of similarity of the open sets in $\mathcal{U} $ is essential.

This is my approach: Given that the Lebesgue measure $\lambda$ is inner regular, pick a compact set $K \subset W$ such that
$$
\lambda(W) - \epsilon < \lambda(K).
$$
Since the open sets\footnote{We take it as given that $C$ is open implies that $x + rC$ is open.} in $\mathcal{U}$ covers $K$ and $K$ is compact there exists a finite number $m$ of them such that $U_i, \dots, U_m$ covers $K$. Order the $U_i$:s such that, if
$$
U_i = x_i + r_iC, \qquad i=1, \dots m
$$
then $r_1 \ge r_1 \ge \dots \ge r_m$.
Now, I have seen a similar Lemma in Rudin's Real and complex analysis (Lemma 7.3) where the sets $U_i$ are open balls $U_i = B(x_i,r_i)$. In that case one may do as follows: to get a disjoint collection of sets we let $i_1=1$, and then we discard every $U_j$ that intersects $U_{i_1}$. Let $U_{i_2}$ be the first remaining $U_j$ (if any exists) and discard the remaining $U_j$ that intersects $U_{i_2}$. Continuing this process gives a a collection of $n$ disjoint sets. Then one may claim that,
$$
\bigcup_{i=1}^m x_i+ r_iC \subset \bigcup_{k=1}^n x_{i_k} + 3r_{i_k}C.
$$
The conclusion then follows (for that Lemma) from
$$
\lambda \left(B(x, 3r)\right) =3^d \lambda \left(B(x, r)\right) 
$$
and subadditivity.

I don't think the exact same argument works here but is there a similar approach to get at least pairwise disjoint sets in the case of this exercise? And further such that (1) holds?


Secondly, what is the meaning of the last sentence in the exercise, "Show by a counterexample that the condition of similarity of the open sets in $\mathcal{U} $ is essential."?

Much grateful for any help provided!
 A: 
I don't think the exact same argument works here but is there a similar approach to get at least pairwise disjoint sets in the case of this exercise? And further such that (1) holds?

The proposed  argument is used to prove the Vitali covering lemma.
As Martin Argerami noted, it needs that $C=x-C$ for some $x\in\Bbb R^d$. Then we can replace it by a set $C’=C-x/2=-C'$ containing the origin of $\Bbb R^d$ and either (provided $C’$ is non-empty) construct Minkowski functional to endow $\Bbb R^d$ with the metric allowing to apply Vitali covering lemma for a metric space or directly providing the required claim:  if $$x_i+r_iC’\cap x_j+r_jC’\ne\varnothing$$ and $r_i\ge r_j$ then  $ x_j+r_jC’ \subset x_i+3r_iC'$. Indeed, let $$x\in x_i+r_iC’\cap x_j+r_jC’$$ be an arbitrary point and $y\in x_j+r_jC’$. Then $$y\in x_j + r_jC' \subset x-r_jC’+ r_jC’\subset x_i+r_iC’-r_jC’+ r_jC’\subset x_i+3r_iC’.$$
But the condition $C=x-C$ was missed and the argument doesn’t work now.
Moreover, the exercise claim is wrong as the following example shows.  Indeed, consider a simplex
$$C=C_d=\{(x_1,\dots,x_d):  x_i>0\mbox{ for each }i\mbox{ and } x_1+\dots+x_d<1\}.$$
We have $\lambda(C_d)=\frac 1{d!}$. Let $\mathcal U=\{x+C: x\in -C\}$. Then $W=\bigcup U=C-C$. Since each member of $\mathcal U$ contains the origin of $\Bbb R^d$, $\mathcal U$ has no disjoint subsets.
Let’s calculate $\lambda(W)$. For each $\delta=(\delta_1,\dots,\delta_d)\in \{-1,1\}^d$ put
$$W_\delta=\{(x_1,\dots, x_d)\in W: \forall i (\delta_ix_i>0) \}.$$
Let $\delta_+=\{1\le i\le d: \delta_i=1\}$, $\delta_-=\{1\le i\le d: \delta_i=1\}$, and $\pi_+$ and $\pi_-$ be the projections of the product $\Bbb R^n=\Bbb R^{\delta_+}\times \Bbb R^{\delta_-}$ into its factors $\Bbb R^{\delta_+}$ and $\Bbb R^{\delta_-}$, respectively. Put $k=|\delta_+|$. It is easy to see that if $1\le k\le d-1$ then $\pi_+(W_\delta)$ is a natural copy of $C_k$,  $\pi_-(W_\delta)$ is a natural copy of $C_{d-k}$, and $W_\delta=\pi_+(W_\delta)\times \pi_-( W_\delta)$. Thus $\lambda(W_\delta)=\frac 1{k!} \tfrac 1{(d-k)!}$.
So we have $$\lambda(W)=\sum\{\lambda(W_\delta): \delta\in \{-1,1\}^d \} =$$
$$\sum\left\{\frac 1{k!} \frac 1{(d-k)!}: \delta\in \{-1,1\}^d \mbox{ and } |\delta_+|=k\right\}=$$
$$\sum_{k=0}^d \frac 1{k!} \frac 1{(d-k)!}\cdot |\{\delta\in \{-1,1\}^d \mbox{ and } |\delta_+|=k\}|=$$
$$\sum_{k=0}^d\frac 1{k!} \frac 1{(d-k)!}{d\choose k}=\frac 1{d!}\sum_{k=0}^d {d\choose k}^2=\frac 1{d!} {2d\choose d}=\lambda(C){2d\choose d}$$
(see here) for the last equality.
Finally, Robbins’ bounds imply that
$$\frac {4^{d}}{\sqrt{\pi d}}\exp\left(-\frac {1}{8d-1}\right)<{2d\choose d}<\frac {4^{d}}{\sqrt{\pi d}}\exp\left(-\frac {1}{8d+1}\right).$$
A: For your last question regarding the statement:

Show by a counterexample that the condition of similarity of the open sets in $\mathcal U$ is essential.

This means (as I understand it) that if $\mathcal U$ simply is a collection of convex open sets, and not necessarily sets of the form $x + rC$ for a fixed convex open set $C$, then this result fails.
Here is one counterexample in $\mathbb R^2$. Let $q \in (0,1)$, and let $\tilde C_0$ be the sector given in polar coordinates by $\{(r,\theta) : 0 < r < 1, \, 0 < \theta < 2\pi q\}$. Let $A_\theta : \mathbb R^2 \to \mathbb R^2$ denote a rotation by $\theta$, let $\alpha > 0$, let $a = -\alpha(\cos\pi q, \sin\pi q) \in \mathbb R^2$, and let $C_\theta = A_\theta\left(a+\tilde C_0\right)$. Then each $C_\theta$ is an open sector with radius $1$ and angle $2\pi q$ containing the origin. If $\mathcal U = \left\{ C_\theta : 0 \leq \theta <2\pi\right\}$, $W = \bigcup_{\theta \in [0,2\pi)} C_\theta$, then we find that $W$ is the open disc of radius $1-\alpha$ centered at the origin. So $\lambda^2(W) = \pi(1-\alpha)^2$, and $\lambda^2(C_\theta) = \pi q$ for every $\theta$. Since no two sets in $\mathcal U$ are disjoint, the statement of the result implies:
$$
\forall \epsilon > 0 : \quad \pi q = \lambda^2(C_\theta) > \frac{1-\epsilon}{9}\lambda^2(W) = \frac{1-\epsilon}9 \pi(1-\alpha)^2.
$$
Since this must hold for every $\epsilon > 0$, this means in particular that $q \geq \frac 1 9 (1-\alpha)^2$. This is false given appropriate choices of $q$ and $\alpha$, e.g. if $q \in (0,\frac 1 9)$ and if $\alpha < 1-3\sqrt q$.
