TITLE: Clarifying the relationship between outer measures, measures and measurable spaces: the converse direction
QUESTION [27 upvotes]: This is related to my measure theory class, but it's not homework. The motivation behind this post is to understand the exact relationship between the three objects mentioned in the title.
Carathéodory's theorem states that given an outer measure $\varphi$ on a space $X$, the set of $\varphi$-measurable sets form a $\sigma$-algebra and $\varphi$ restricted to the measurable sets is a countably additive complete measure. I am wondering if this admits a converse:

Q1. Given a complete measure space $(X, \Sigma, \mu)$, does there always exist an outer measure $\varphi$ satisfying the following conditions?

A set $E \subseteq X$ is $\varphi$-measurable iff $E \in \Sigma$,
$\varphi$ restricted to $\Sigma$ is $\mu$.


A related question is:

Q2. Given a $\sigma$-algebra $\Sigma$ on $X$, can I always define a complete measure $\mu$ on the measurable space $(X, \Sigma)$?

To pinpoint the source of my confusion, I feel that the word measure is used in several related, but seemingly different, ways; e.g., "measure space", "measurable space" and "outer measure". If the answers to both the questions are in the affirmative, then it would become clearer to me that these different usages are very related after all.

REPLY [5 votes]: I am posting this because I think it's quite relevant to my question, even if it doesn't directly address it. This is inspired by t.b.'s answer and Niels' comment, and it hopefully complements them both. 
Interpreting my second question broadly, we want to understand whether a measurable space admits a measure (with additional restrictions). Of course, the zero measure trivially works for all measurable spaces $(X, \Sigma)$. Slightly more interestingly,

For an arbitrary $x_0 \in X$, the measure
$$
\mu_{x_0}(A) = \begin{cases} 1, &x_0 \in A, \\ 0, &x_0 \notin A, \end{cases}
$$
is a probability measure (consequently also $\sigma$-finite).
The measure $$
\mu(A) = \begin{cases}
\infty, &A \neq \emptyset,
\\ 0, &A = \emptyset,
\end{cases}
$$
suggested by Niels is a complete measure, but it fails to be $\sigma$-finite (and fails quite badly too). Of course, this technically answers my second question.

However, one could further ask whether every measurable space admits a measure that is both complete and sigma-finite. This combination of properties is attractive because this is precisely the sufficient condition given by t.b.'s answer for the measure to be identical to that given by the associated outer measure. Unfortunately, we show that this is too much to hope for in general. 

As a counter-example, take $I$ to be an uncountable index set, and $\{ X_\alpha \}_{\alpha \in I}$ to be a collection of disjoint 2-element sets. Define $X = \bigcup_{\alpha \in I} X_\alpha$, and consider the $\sigma$-algebra
$$
\Sigma = \{ A \subseteq X \,:\, \forall \alpha \in I, \text{ either } A \cap X_{\alpha} = \emptyset \text{ or } X_\alpha \subseteq A \}.
$$ 
Suppose $\mu$ is a complete measure on $(X, \Sigma)$. Since none of the proper subsets of $X_\alpha$ are in $\Sigma$, completeness requires that $\mu(X_{\alpha}) > 0$ for all $\alpha \in I$. 
Now, any countable partition of $X$ into measurable sets must be of the form
$$
\left\{ \bigcup_{\alpha \in I_n} X_{\alpha} \right\}_{n \in \mathbb N}
$$
where $\{I_n\}_{n \in \mathbb N}$ forms a partition of $I$. Since $I$ is uncountable, it follows that there exists $n$ such that $J := I_n$ is uncountable. We show that $\bigcup_{\alpha \in J} X_\alpha$ has infinite measure. (This will complete the proof that the measure $\mu$ is not $\sigma$-finite.) 
The remaining parts of the proof also follow standard ideas. Notice that, by the Archimedean property of the reals, we can write $J$ as a countable union
$$
J = \bigcup_{m \in \mathbb N} \left \{ \alpha \in J \,:\, \mu(X_\alpha) \geq \frac{1}{m} \right\}.
$$
Since $J$ is uncountable, it follows that there exists $m \in \mathbb N$ such that 
$$
J_m := \left \{ \alpha \in J \,:\, \mu(X_\alpha) \geq \frac{1}{m} \right\}
$$
is infinite (in fact, uncountable as well). Since each $\alpha \in J_m$ is such that $X_\alpha$ accounts for at least $1/m$ measure, and all the $X_\alpha$'s are disjoint, it follows that the total measure of the set $$
\bigcup_{\alpha \in J_m} X_\alpha
$$
is infinite. Finally, since $J \supseteq J_m$, we are done by the monotonicity of $\mu$.