I would really like to show that the following is true.
"Suppose that $X$ is a set and $\theta$ is an outer measure on $X$, and let $\mu$ be the measure on $X$ defined by Carathéodory's method. Then if $\theta E = 0$, then $\mu$ measures $E$."
I'm not exactly sure how $E$ is defined, which could be the problem, but the question goes onto ask me to deduce that if $E \subseteq X$ is $\mu$-negligible iff $\theta E = 0$, so I assume that this is the same $E$ as in the statement above.
I have been using the following definitions and theorems.
Definition (sigma algebra) Let $X$ be a set. A *$\sigma$-algebra of subsets of $X$ is a family $\Sigma$ of subsets of $X$ such that
(i) $\emptyset \in \Sigma$;
(ii) for every sequence $\left< E_n \right>_{n \in \mathbb{N}}$ in $\Sigma$, its union $\bigcup _{n \in \mathbb{N}} E_n$ belongs to $\Sigma$.
Definition (measure space) A measure space is a triple $(X, \Sigma, \mu)$ where
(i) $X$ is a set;
(ii) $\Sigma$ is a $\sigma$-algebra of subsets of X;
(iii) $\mu : \Sigma \rightarrow [0, \infty]$ is a function such that
(a) $\mu \emptyset = 0 $;
(b) if $\left
In this context, members of $\Sigma$ are called measurable sets, and $\mu$ is called a measure on $X$.
Definition (outer measure) Let $X$ be a set. An outer measure on $X$ is a function $\theta : \mathcal{P}X \rightarrow \left[ 0 , \infty \right]$ such that
(i) $\theta \emptyset = 0$,
(ii) if $A \subseteq B \subseteq X$ then $\theta A \leq \theta B$,
(iii) for every sequence $\left< A_n \right> _{n \in \mathbb{N}}$ of subsets of $X$, $\theta \left( \bigcup _{n \in \mathbb{N}} A_n \right) \leq \sum_{n=1}^{\infty} \theta A_n$.
Carothéodory's Method: Theorem Let X be a set and $\theta$ an outer measure on $X$. Set
$ \Sigma := \left\{ E \subseteq X : \theta A = \theta \left(A \cap E \right) + \theta \left( A \setminus E \right), \forall A \subseteq X \right\}. $
Then $\Sigma$ is a $\sigma$ algebra of subsets of $X$. Define $\mu : \Sigma \rightarrow \left[0, \infty \right]$ by writing $\mu E = \theta E$ for $E \in \Sigma$; then $\left( X, \Sigma, \mu \right)$ is a measure space.
I think I've followed the proof I have for the above Carothéodory's Method, and I suspect that the proof of the statement in question shall follow it, but proving instead that $\left(A, \Sigma, \mu \right)$ is a measure space. Perhaps the proof is trivial? I just can't see it.