# Carathéodory's method gives a complete measure

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 $\left<E_n \right>_{n \in \mathbb{N}}$ is a disjoint sequence in $\Sigma$, then $\mu \left( \bigcup _{n \in \mathbb{N} } E_n \right) = \sum_{n=1}^{\infty} \mu E_n$.

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.

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I never heard the expression "$\mu$ measures $A$" used this way, but from context: Isn't the intention that you have to show: for any subset $A \subset X$ and $\Sigma$ as in Carathéodory's method, $\theta A = 0$ implies $A \in \Sigma$? –  t.b. Jul 21 '11 at 0:15
I usually see $\theta$-measurable (referencing the outer measure). –  Dylan Moreland Jul 21 '11 at 0:23
Also, $A$ could be any subset of $X$. You need only show (why?) that for each $E \subset X$ we have $\theta(E \cap A) + \theta(E \setminus A) \leqq \theta(E)$. Use that $\theta$ is subadditive! –  Dylan Moreland Jul 21 '11 at 0:40
True that. $A$ is usually the measurable guy and $A$ the random arbitrary bad guy. But still, +1 for adding definitions and doing your best to be clear. –  Patrick Da Silva Jul 21 '11 at 0:46
Just so you now: the fancy way of stating your claim is "Carathéodory's method gives a complete measure". See wiki: en.wikipedia.org/wiki/Complete_measure –  Mark Jul 21 '11 at 0:59

EDIT : Ok so, we want to show that if $\theta E =0$, we have that $E \in \Sigma$. Now since outer measures are subadditive, we have $$\theta A \le \theta(A \cap E) + \theta (A \backslash E)$$ thus it remains to prove that $$\theta A \ge \theta(A \cap E) + \theta( A \backslash E).$$ Now $0 \le \theta(A \cap E) \le \theta E = 0$ since $\theta$ is monotonic, and $$\theta(A) \ge \theta(A \backslash E)$$ again since $\theta$ is monotonic, which means $E \in \Sigma$. Now I don't think I'm wrong, correct me if I am.

Hope that helps,

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I didn't downvote, but there are issues: in the second display, you want $0 \leqq \theta(A \setminus E) \leqq \theta(A)$, which could well be positive. But that's okay! –  Dylan Moreland Jul 21 '11 at 1:03
I'm not sure why you got a downvote! Certainly, I am most grateful for your post. I am a little confused, though why it should be that $\theta \left( A \setminus E \right) \leq \theta E$, since it is surely not necessarily the case that $A \setminus E \subseteq E$, right? I wonder what I'm missing? –  Harry Williams Jul 21 '11 at 1:13
Yeah Harry, that's the whole thing : my notation before was making sense but it wasn't proving what I wanted, and now it's screwed because I fixed the notation. Hence I stare at my answer and try to find what I could do to fix this. Gimme a few minutes. –  Patrick Da Silva Jul 21 '11 at 1:14
Finally. I'm more confident now that my answer is fine. Please check it XD it'll make it worth the trouble. –  Patrick Da Silva Jul 21 '11 at 1:30
Your solution is all the more impressive therefore! –  Harry Williams Jul 21 '11 at 1:38

$\theta(A) = 0$

$\forall B\in 2^X$

$A\cap B \subseteq A \;$ and $\; A-B \subseteq A$
$0\leq \theta(A\cap B) \leq \theta(A) = 0 \;$ and $\; 0\leq \theta(A-B) \leq \theta(A) = 0$
$\theta(A) = 0 = 0+0 = \theta(A\cap B)+\theta(A-B)$

$A\in \Sigma$

$A$ is $\mu$-null
iff
$(\exists B\in \Sigma)(A\subseteq B \text{ and } \mu(B) = 0)$
iff
$(\exists B\in \Sigma)(A\subseteq B \text{ and } \theta(B) = 0)$
iff
$(\exists B\in 2^X)(A\subseteq B \text{ and } \theta(B) = 0)$
iff
$(\exists B\in 2^X)(A\subseteq B \text{ and } 0\leq \theta(A) \leq \theta(B) = 0)$
iff
$(\exists B\in 2^X)(A\subseteq A \text{ and } 0\leq \theta(A) \leq \theta(A) = 0)$
iff
$A\subseteq A \text{ and } 0\leq \theta(A) \leq \theta(A) = 0$
iff
$A\subseteq A \text{ and } \theta(A) = 0$
iff
$\theta(A) = 0$

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While this is certainly all correct, this is a model example of how not to write mathematics. We are not robots! –  t.b. Jul 21 '11 at 0:45
Yeah.... The guy's just confusing. How about you use words a little. Speak in english too. –  Patrick Da Silva Jul 21 '11 at 0:48
While this is certainly correct I doubt it will clarify the issue asked about her. I don't know if Ricky deserves such an insane amount of down-voting, though! –  james Jul 21 '11 at 2:56
The proof is wrong in top of that, to prove that $A \in \Sigma$, it must be that for all sets $E \subseteq X$, $\theta(E) = \theta(E \cap A) + \theta(E \backslash A)$, which makes it even more worth downvoting, because there's no follow-up on the answer, it's wrong and it's un-readable. Sorry. I would be willing to remove my down-vote if there would be some editing or signs of work to catch up. –  Patrick Da Silva Jul 21 '11 at 5:53