Given measurable sets $E_k$, find a measurable set $A$ with $0<\mu(E_k\cap A)<\mu(E_k)$

Question

Let $\mu$ be the Lebesgue measure on $\mathbb{R}^n$ and $E_k\subseteq\mathbb{R}^n,k\in\mathbb{N}$ be measurable sets each having positive Lebesgue measure.

How to find a measurable set $A\subseteq\mathbb{R}^n$ so that $0<\mu(E_k\cap A)<\mu(E_k)$ for every $k\in\mathbb{N}$?

Answer 1

It should be understood that "positive measure" sets are Lebesgue measurable. The set $\mathbb N$ of natural numbers starts with $0$.

Lemma 1. Given positive measure sets $E_k$ $(k\in\mathbb N)$, we can find a positive measure set $D\subseteq E_0$ such that $\mu(E_j\setminus D)\gt0$ for all $j\ge1$.

Proof. We may assume that $\mu(E_0)\lt\infty$. For each integer $j\ge1$ find a positive measure set $F_j\subseteq E_j$ with $\mu(F_j)\le\mu(E_0)/2^{j+1}$, and let $F=\bigcup_{j\ge1}F_j$, so that $\mu(F)\lt\mu(E_0)$. Let $D=E_0\setminus F$.

Lemma 2. Given positive measure sets $E_k$ $(k\in\mathbb N)$, we can find pairwise disjoint positive measure sets $D_k\subseteq E_k$ $(k\in\mathbb N)$.

Proof. By Lemma 1, we can find a positive measure set $D_0\subseteq E_0$ such that $\mu(E_j\setminus D_0)\gt0$ for all $j\ge1$.Next, we can find a positive measure set $D_1\subseteq E_1\setminus D_0$ such that $\mu(E_j\setminus(D_0\cup D_1))\gt0$ for all $j\ge2$. Continuing in this way, we get a sequence of positive measure sets $D_k$ $(k\in\mathbb N)$ such that $D_k\subseteq E_k\setminus(D_0\cup\cdots\cup D_{k-1})$ and $\mu(E_j\setminus(D_0\cup\cdots\cup D_k))\gt0$ for all $j\ge k+1$.

Theorem. Given positive measure sets $E_k$ $(k\in\mathbb N)$, we can find a measurable set $A$ such that $0\lt\mu(E_k\cap A)\lt\mu(E_k)$ for each $k\in\mathbb N$.

Proof. By Lemma 2 we can find pairwise disjoint positive measure sets $D_k\subseteq E_k$ ($k\in\mathbb N$); moreover, we can assume that $\mu\left(\bigcup_{k\in\mathbb N}D_k\right)\lt\infty$. For each $k\in\mathbb N$ choose a measurable set $A_k\subseteq D_k$ with $0\lt\mu(A_k)\lt\mu(D_k)$, and let $A=\bigcup_{k\in\mathbb N}A_k$. Now we have $$E_k\cap A=A_k\cup[(E_k\setminus D_k)\cap A],$$ whence $$\mu(E_k\cap A)\ge\mu(A_k)\gt0$$ and $$\mu(E_k\cap A)\le\mu(A)\le\mu\left(\bigcup_{k\in\mathbb N}D_k\right)\lt\infty.$$ If $\mu(E_k)=\infty$ we're done; if $\mu(E_k)\lt\infty$, then $$\mu(E_k\cap A)\le\mu(A_k)+\mu(E_k\setminus D_k)\lt\mu(D_k)+\mu(E_k\setminus D_k)=\mu(E_k).$$

Answer 2

Start with the case that $ \{E_k\} $ are pairwise disjoint and have positive measure. We can then find by inner regularity we can find sets $ \{A_k\} $ so that $ A_k\subset E_k $ and $ 0<\mu(A_k) < \mu(E_k)$. Then $A = \bigcup A_k$ has the desired properties from the disjointness and $ A $ is the union of measurable sets and is thus also measurable. For the general case you just need to transform $ \{ E_k\}$ into a disjoint collection $ \{F_k\} $ by using $F_1 = E_1$ and $F_k = E_k\backslash (F_{1}\cup \ldots F_{k-1})$. The remaining details are straightforward.