I was studying Thomas Jech's proof of the above in his book "The Axiom of Choice" (page 178). The proof starts as follows:
"A non-measurable set has different inner and outer measures. In fact, given a non-measurable set, one can construct a subset of the interval $[0, 1]$ with inner measure $0$ and outer measure $1$. Thus to show that every set is measurable, it suffices to show that for every $X\subseteq [0,1]$, either $X$ contains a subset $U$ of positive measure, or $X$ is disjoint from a set $U\subseteq[0,1]$ of positive measure."
My questions are:
- How can one can construct a subset of the interval $[0, 1]$ with inner measure $0$ and outer measure $1$ and why does this help us here?
- Why does it suffices to just show this? Why will this imply every subset of $\mathbb{R}$ is Lebesgue-measurable?
The definition of Lebesgue-measure Jech's gives is as follows:
The standard way of defining the Lebesgue measure is to define first the outer measure $\lambda^{*}(X)$ of a set $X \subseteq \mathbb{R}^{n}$ as the infimum of all possible sums $\sum\left\{ v(I_k) : k \in \mathbb{N} \right\}$ where $\left\{ I_k : k \in \mathbb{N} \right\}$ is a collection of $n$-dimensional intervals such that $X \subseteq \bigcup_{k=0}^{\infty} I_k$, and $v(I)$ denotes the volume of $I$. For each $X$, $\lambda^{*}(X) \geq 0$ and possibly $= \infty$. A set $X$ is null if $\lambda^{*}(X)=0$.
A set $A \subseteq \mathbb{R}^{n}$ is Lebesgue measurable if for each $X \subseteq \mathbb{R}^{n}$, \begin{equation*} \lambda^{*}(X)=\lambda^{*}(X \cap A) + \lambda^{*}(X \enspace \backslash \enspace A) \end{equation*} For a measurable set $A$, we write $\lambda(A)$ instead of $\lambda^{*}(A)$ and call $\lambda(A)$ the Lebesgue measure of $A$.
Thanks!
