# Open and compact in Lebesgue measure

Let $\mathcal{F}$ an algebra of sets and let $\mu_{0}$ a finite and $\sigma$- additive measure.

We denote with $\mathcal{F_\sigma}$ the family all countable unions of sets of $\mathcal{F}$ and with $\mathcal{F_\delta}$ the family all countable intersection of sets of $\mathcal{F}$.

We extend $\mu_{0}$ to two family $\mathcal{F_\sigma}$ and $\mathcal{F}_{\delta}$ in the following way:

if $A\in\mathcal{F_{\sigma}}$ \begin{equation} \mu_{1}(A)=\sup\{\mu_{0}(A'), A'\subseteq A,\; A'\in\mathcal{F}\} \end{equation} else if $B\in\mathcal{F_\delta}$ \begin{equation} \mu_{2}(B)=\inf\{\mu_{0}(B'), B'\supseteq B,\; B'\in\mathcal{F}\}. \end{equation}

Definition 1 A set $M\subset X$ is measurable if for all $\epsilon>0$, exist $A\in\mathcal{F_{\sigma}}$ and $B\in\mathcal{F_\delta}$ such that $B\subset M\subset A$ and such that \begin{equation} \mu_{1}(A)-\mu_{2}(B)<\epsilon. \end{equation}

Another definition

Definition 2 A bounded set $M$ is Lebesgue-measurable if, taken $k\in\mathbb{N}$ such that $[-k,k]^{n}\supset M$, $M$ is measurable according to Definition 1, where $\mu_{0}$ is Peano-Jordan measure, on the algebra of PJ-measurable in $[-k,k]^{n}$. If $M$ is not bounded, $M$ is measurable if $M\cap Q$ is measurable, where Q varies in the set of "squares" of $\mathbb{R^{n}}$.

Let $\lambda$ is the Lebesgue measure and $m$ Peano-Jordan measure.

Theorem

Let $M\subset \mathbb{R^n}$ a set. Then $M$ is Lebesgue measurable iif for all $\epsilon>0$ exist a compact $C_{\epsilon}$ and an open $A_{\epsilon}$ such that $C_{\epsilon}\subset M \subset A_{\epsilon}$ and such that $\lambda(A_{\epsilon})-\lambda(C_{\epsilon})<\epsilon$.

We denote with $\mathcal{R}$ the set of Peano-Jordan measurable.

$M$ bounded.

Let $Q\supset M$. Then if $M$ is measurable exist $\{A_{n}\}\subset \mathcal{R}$ such that $M\subset\cup_n A_{n}$ and $m(\cup A_{n})-m(M)<\epsilon/4$.

we can assume that it exists a open multi-interval $P_n$ such that $P_n\supset A_n$ and $\lambda(P_n)-\lambda(A_n)<\frac{\epsilon}{2^{n+1}}$.

Now, is simple see

\begin{equation} \lambda(\cup P_n)-\lambda(M)<\epsilon. \end{equation}

Then, $A=\cup_n P_n$ it is the open sought.

For the compact, suppose that $Q$ is open. Let's build an open one $B\subset Q$ such that $B\supset Q\setminus M$ and such that $\lambda(B)-\lambda(Q\setminus M)<\epsilon$.

Now, just consider $C=cl(Q)\setminus B$.

How can I prove the theorem if M is not bounded?

Thanks!

• You could replace "compact" by "closed" and change $\lambda(A_\epsilon) - \lambda(C_\epsilon)$ to $\lambda(A_\epsilon \setminus C_\epsilon)$. – Umberto P. May 22 '18 at 15:26