real analysis - Compact $Ksubset A$ such that $lambda(K) = lambda(A) / 2$

Let $A\subset \mathbb{R}$ be a (Lebesgue) measurable set of finite measure. Using the fact that the function $f:\mathbb{R}\rightarrow \mathbb{R}$, $$f(x)=\lambda(A\cap [-x,x])$$
is continuous, we can find a bounded subset $K\subset A$ such that $\lambda(K) = \lambda(A) / 2$.

Is it possible to choose $K$ to be compact as well?

($\lambda$ denotes the Lebesgue measure)

Answer

Assume $\lambda(A)>0$ (otherwise just take $K=\emptyset$). By the inner regularity of Lebesgue measure, there is a compact set $K_0 \subseteq A$ with $\lambda(K_0)>\lambda(A)/2$. Define
$$g(x)=\lambda(K_0 \cap [-x,x])$$
By the Intermediate Value Theorem, there exists an $x$ such that $g(x)=\lambda(A)/2$, and then $K:=K_0 \cap [-x,x]$ will be a compact set satisfying $\lambda(K)=\lambda(A)/2$, as desired.