I'm brushing up on some measure theory for quals, and would like some input on the following:
Proposition. $E \subseteq \mathbb{R}^d$ measurable iff every box (cartesian product of closed intervals) $Q$ we have
$$m(Q) = m^{\ast}(Q \cap E) + m^{\ast}(Q \setminus E).$$
I recall some equivalent definitions of Lebesgue measurable:
Equivalent Statements of Measurable:
- For all $\epsilon > 0$ there exists an open $U \supseteq E$ such that $m^{\ast}(U \setminus E) \leq \epsilon$.
- A set $E \subset \mathbb{R}^d$ is measurable iff for each $\epsilon > 0$ there exists a closed set $F \subset E$ such that $m^{\ast}(E \setminus F) < \epsilon$.
- Carathéodory's Theorem. $E$ measurable iff for all $X$ one has $m^{\ast}(X) = m^{\ast}(X \cap E) + m^{\ast}(X \setminus E)$.
- $E = G \setminus H$ where $G$ is a $G_{\delta}$-set and $m(H) = 0$.
- $E = F \cup H$ where $F$ is a $F_{\sigma}$-set and $m(H) = 0$.
My proof:
$\Rightarrow$: Suppose $E$ measurable. Since every box (finite cartesian product of close intervals) in $\mathbb{R}^d$ is measurable we have that
$$Q = (Q \cap E) \cup (Q \setminus E),$$
disjoint and measurable. Applying countable additivity gives the result.
$\Leftarrow$: Suppose the equation proposed holds. Then for any $k \in \mathbb{N}$, there exists a box $Q_k$ such that $E_k = Q_k \cap E$. Hence, $E \cap Q_k \subseteq E$, and $E = \cup_{k=1}^{\infty} (E \cap Q_k)$. Furthermore, choose an $\epsilon > 0$ such that there exists and open set $U$ with $E_k \subseteq U$. Then
$$m^{\ast}(E_k) \leq m(U) \leq m^{\ast}(E_k) + \epsilon.$$
WLOG, we can replace $U$ with $U \cap Q_k$. Then $U \subseteq Q_k$. Hence,
$$ \begin{align} m^{\ast}(E_k) + m^{\ast}(U \setminus E_k) &= m^{\ast}(U \cap E_k) + m^{\ast}(U \setminus E_k) \\ &= m^{\ast}(U \cap E) + m^{\ast}(U \setminus E) \\ &= m(U) \\ &\leq m^{\ast}(E_k) + \epsilon. \end{align} $$
$m^{\ast}(E_k)$ finite, so subtract from both sides and we are done.
Not too sure about $\Leftarrow$, particularly if the $E_k$ construction is valid. I know that in Carathéodory's Theorem, a similar construction is used but with balls $B_k(0)$, which gave me the idea.