Is the Lebesgue measure on $[a,b]$ $(\subset \mathbb{R})$ perfect?

Question

We know, that $[a,b]$ is perfect when the sigma algebra is that of Borel sets. Does it hold true when the sigma algebra is that of Lebesgue measurable sets?

Definition:A measure space $(X, Σ, μ)$ is said to be perfect if, for every $Σ$-measurable function $f : X → \mathbb{R}$ and every $A ⊆ \mathbb{R}$ with $f^{−1}(A) ∈ Σ$, there exist Borel subsets $A_1$ and $A_2$ of $\mathbb{R}$ such that

${\displaystyle A_{1}\subseteq A\subseteq A_{2}{\mbox{ and }}\mu {\big (}f^{-1}(A_{2}\setminus A_{1}){\big )}=0.}$

Any help would be much appreciated. Thank you.

Answer 1

Your question is confusing. The notion of a perfect set is a topological one. However, in the title you're asking about the perfectness of the Lebesgue measure ($\lambda$) on $[a,b]\subset \mathbb{R}$.

First, $\lambda$ is a perfect measure on $\mathcal{B}_{[a,b]}$ ($\because$ it is inner regular). Now using the following result (Proposition 7.5.4(ii) here), it is clear that $\lambda$ is a perfect measure on the completion of $\mathcal{B}_{[a,b]}$:

Proposition. If a measure $\mu$ on a $\sigma$-algebra $\mathcal{S}$ is perfect, then it's restriction to any set $E\in \mathcal{S}_{\mu}$ (the completion of $\mathcal{S}$ w.r.t. $\mu$) equipped with the trace of an arbitrary sub-$\sigma$-algebra in $\mathcal{S}_{\mu}$ is a perfect measure.