is there any measurable set which is not open,close or Borel?

Question

is there any measurable set which is not open,close or Borel? if yes please gimme an example.

Answer 1

Here I assume Real line $\Bbb R$ with Lebesgue measure.

There exists a subset $S\in \Bbb R$ which is Lebesgue measurable but not a Borel set , since cardinality of set of all Borel set is $\mathscr c$ but cardinality of set of all Lebesgue measurable set is $2^{\mathscr c}$ , where $\mathscr c $ is the cardinality of $\Bbb R$.

If $\phi$ is the cantor function then $\psi :[0,1]\rightarrow [0,2]$ defined by $\psi(x)= x+\phi (x)$ is a homeomorphism. Let $D$ be a non-measurable subset (any measurable subset of positive measure contains a non-measurable subset due to Axiom of Choice) of $\psi (C)$ where $C$ is the cantor set . Then $\psi^{-1} (D)$ is a subset of $C$ and hence is a measurable set of Lebesgue measure $0$ (Lebesgue measure is complete). However since $\psi^{-1} (D)$ is the image under a homeomorphism of a non-Borel set $D$ (as $D$ is not measurable), $\psi^{-1}(D)$ is not a Borel set.

The function $\phi:C\rightarrow \Bbb R$ defined by $\phi(x)=\sum_{n=1}^{\infty} \frac{a_n}{2^{n+1}}, x=\sum_{n=1}^{\infty} \frac{a_n}{3^n}$ is a monotone increasing continuous function which is called cantor function.