0
$\begingroup$

Show that every open and closed subsets of $\mathbb{R}$ are Lebesgue measurable. The measure $m$ defined on $\mathscr{L}$ by $m(A) = m^*(A)$ is called the Lebesgue measure on $\mathbb{R}$.

1.) $m : \mathscr{E} \rightarrow [0, \infty)$ by $m(\emptyset) = 0$ and if $\bigcup_{n=1}^N I_n \in \mathscr{E}$, a disjoint union of intervals, then $$m\left(\bigcup_{n=1}^N I_n\right) = \sum_{n=1}^N \ell(I_n).$$ $\mathscr{E}$ is the collection of subsets of $\mathbb{R}$ which are finite unions of disjoint intervals.

2.) $\mathscr{L}$ is the collection of subsets in $\mathbb{R}$ which can be written as a countable union of sets in $\bar{\mathscr{E}}$ where $\bar{\mathscr{E}}$ is the collection of subsets $A \subseteq \mathbb{R}$ for which there exist a sequence $(A_n)$ in $\mathscr{E}$ with $d(A_n , A) \rightarrow 0$ as $n \rightarrow \infty$. Note that $d$ is a semi-metric defined as $d : \mathscr{P}(\mathbb{R}) \times \mathscr{P}(\mathbb{R}) \rightarrow [0, \infty]$ by $d(A, B) = m^*(A \triangle B)$ (Symmetric difference).

I am new to this material and confused on how to prove this. For any open subset $O$ of $\mathbb{R}$, we can write $O$ as a countable union of disjoint open intervals so $O = \bigcup_{n=1}^\infty I_n$. Then, since $m^*$ is countably subadditive we have $$m^*(O) \leq \sum_{n=1}^\infty m^*(I_n)$$ where $m^*(A) = \inf\left\{\sum_{n=1}^\infty \ell(I_n) : I_n \ \text{is an open interval and} \ A \subseteq \bigcup_{n=1}^\infty I_n\right\}$. Am I on the right track? Any help will be appreciated.

  • 0
    Do you know the concept of $\sigma$-algebra? – 2017-02-27
  • 0
    What definition of "Lebesgue measurable" are you using? – 2017-02-27
  • 0
    Ok I updated everything. It is a lot of definitions. I might be getting confused because of all these definitions and collection of sets. – 2017-02-27
  • 0
    The proof is standard. You can see the book of Royden. The idea is that the collection $\mathcal{M}$ of measurable sets is a $\sigma$-algebra. Next, you have to show that for every $a\in\Bbb R$, the set $(a,+\infty)$ is measurable. This implies that every open interval $(a,b)$ is measurable. Hence, every open set must be measurable. Measurability of a closed set is easy to show. – 2017-02-27

0 Answers 0