Is the Lebesgue Measure on $[a,b]$ perfect?

Question

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 $f^{−1}(A) \in Σ$, there exist Borel subsets $A_1$ and $A_2$ of $\mathbb{R}$ such that

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

$\bf\underline{A \ result}$ $\textit{ (Given in the wiki page):}$

If $X$ is any metric space and $μ$ is an inner regular (or tight) measure on $X$, then $(X, B_X, μ)$ is a perfect measure space, where $B_X$ denotes the Borel $σ$-algebra on $X$.

Inner Regular Measure: Lebesgue measure on $\mathbb{R}$ is inner-regular.

From the information given above, we can conclude that $\displaystyle(\mathbb{R},B_{\mathbb{R}}, \lambda)$ is a perfect measure space, $\lambda$ being the Lebesgue Measure.

Again, A measure is called inner regular if every measurable set is inner regular. Any compact subset $G$ of $\mathbb{R}$ is Lebesgue measurable.

Therefore, $(G,B_G, \lambda)$ is a perfect measure space, where $G$ is any compact subset of $\mathbb{R}$, $B_G$ is the Borel $\sigma$-algebra on $G$ and $\lambda$ is the Lebesgue Measure.

Is this reasoning correct? Kindly verify.

Answer 1

You are confusing two of the related, but different concepts: your argument shows that $G$ is an inner regular set in $\mathbb{R}$, but this is not the same as $G$ (with the inherited structures) being an inner regular measure space. Therefore, your argument is not quite correct, but a proper argument proceeds along the same lines.

Let's play a game of unraveling definitions (I'm following Wikipedia for the conventions):

Definition: A measure space $(X,\Sigma,\mu)$ is perfect, if, for every $\Sigma$-measurable function $f\colon X\rightarrow\mathbb{R}$ and every $A\subseteq\mathbb{R}$ with $f^{-1}(A)\in\Sigma$, there exist Borel subsets $A_1,A_2\subseteq\mathbb{R}$ such that $A_1\subseteq A\subseteq A_2$ and $\mu(f^{-1}(A_2\setminus A_1)=0$.

Definition: Let $(X,\tau)$ be a topological space and $(X,\Sigma,\mu)$ a measure space. A measurable set $A\subseteq X$ is said to be inner regular (with respect to $\mu$) if $\mu(A)=\sup\{\mu(K)\colon\ K\subseteq A,K\text{ compact and measurable}\}$.

Definition: Let $(X,\tau)$ be a Hausdorff topological space and let $\Sigma$ be a $\sigma$-algebra on $X$ containing the topology. Then a measure $\mu$ on $(X,\Sigma)$ is called inner regular if every measurable set $A\subseteq X$ is inner regular with respect to $\mu$.

Theorem: If $X$ is a metric space, $\mathcal{B}_X$ is the Borel-algebra on $X$ and $\mu$ is an inner regular measure on $(X,\mathcal{B}_X)$, then $(X,\mathcal{B}_X,\mu)$ is perfect.

Now, we will be able to prove the following:

Proposition: Let $G\subseteq\mathbb{R}$ be Borel-measurable and equip it with the subspace topology. Then $(G,\mathcal{B}_G,\lambda\mid_{\mathcal{B}_G})$ is a perfect measure space.

Proof: Note that $G$ is Hausdorff and the Borel-algebra on $G$ contains the topology by definition. By the Theorem, it suffices to show that $\lambda\mid_{\mathcal{B}_G}$ is an inner regular measure on $G$. To that end, let $A\subseteq G$ be a measurable subset. Then $A$ is a measurable subset of $\mathbb{R}$, because $\mathcal{B}_G\subset\mathcal{B}_{\mathbb{R}}$ (Why? This requires the definition of subspace topology and uses that $G$ be measurable). By the inner regularity of the Lebesgue measure on $\mathbb{R}$, we have $\mu(A)=\sup\{\mu(K)\colon K\subseteq A,K\text{ compact subset of }\mathbb{R}\}$ (the condition of measurability is superfluous here, because any compact subset is Borel-measurable). But any of these $K$ is contained in $G$ and since $G$ is equipped with the subspace topology, $K\subseteq G$ is compact in $G$ iff it is compact in $Y$ (which is to say the subspace topology on $K$ inherited from $G$, which inherited its topology from $\mathbb{R}$, is the same as the subspace topology $K$ inherits directly from $\mathbb{R}$). Furthermore, $\lambda\mid_{\mathcal{B}_G}(K)=\lambda(K)$ for $K\subseteq G$ by definition of the restriction. It follows that $$\lambda\mid_{\mathcal{B}_G}(A)=\lambda(A)=\sup\{\lambda(K)\colon\ K\subseteq A,\,K\text{ compact in }\mathbb{R}\}\\=\sup\{\lambda\mid_{\mathcal{B}_G}(K)\colon K\subseteq A,\,K\text{ compact in }G\}.$$ Note that any compact subset of $G$ is Borel-measurable, so this means that $A$ is an inner regular subset of $G$. Since $A$ was arbitrary, $\lambda\mid_{\mathcal{B}_G}$ is an inner regular measure on $(G,\mathcal{B}_G)$.