Intersection of Borel sets with positive Lebesgue-Borel measure

Question

Let $A \in \mathfrak{B}(\mathbb{R})$ be a Borel set with $\lambda(A) > 0.$ Are there $B, C \in \mathfrak{B}(\mathbb{R})$ with $ B\cap C =\emptyset, B\cup C = A$ and $\lambda(B), \lambda(C) > 0?$

Thank you.

Answer 1

If $\lambda (A\cap (-\infty,0])>0$ and $\lambda (A\cap (0,\infty))>0$ then we are finished.

If $\lambda (A\cap (0,\infty))=0$, replace $A$ by $-A$, so we can assume that $\lambda (A\cap (-\infty,0))=0$.

Let $f(x) = \lambda ( A \cap (-\infty, x])$. $f$ is continuous, non decreasing, $\lim_{x \to -\infty} f(x) = 0$, $\lim_{x \to \infty} f(x) = \lambda(A)$, so there is some $x$ such that $0<f(x) < \lambda(A)$.

Choose $ B= A \cap (-\infty, x]$, $C = A \setminus B$.

Answer 2

$$\text {Let} \quad U = \{q\in Q:\; 0= \lambda (A\cap (-\infty,q)\;)\}.$$ $$\text {Let }\quad V=\{q\in Q:\; 0=\lambda (A\cap (q,\infty)\;)\}.$$ In $[-\infty, \infty],$ let $u_1=\sup U$ and $v_1=\inf V.$ We have $u_1<v_1.$ Then we have

$$\bullet \quad \forall q\in Q\cap (u_1,v_1) \; [\;\lambda (A\cap (-\infty,q))>0<\lambda (A\cap (q,\infty))\;].$$

Note (1): If $u_1\geq v_1$ then $Q$ \ $\{u_1,v_1\} \subset ( U\cup V ),$ but since $Q$ is countable and dense in the reals, this would imply $\lambda (A)=0.$

Note (2): Since $u_1<v_1,$ the line $\bullet$ follows from the def'ns of $u_1$ and $v_1$.