I'm only going to address question 1 in some generality, since the infinite measure $\mu(E) = \infty$ for $\emptyset \neq E \in \Sigma$ gives a complete measure on $\Sigma$ and thus an obvious answer to question 2 (as was noted by Niels Diepeveen).

Most of the things I'm saying below can be found or extracted from Fremlin's *Measure theory*, volumes 1 and 2, parts even verbatim, up to minor modifications of notation.

Let me say right away that the situation is somewhat subtle. The answer I'm going to give boils down to:

Every measure $\mu$ is induced by an outer measure $\mu^{\ast}$ canonically obtained from $\mu$, but in general the $\sigma$-algebra of $\mu^{\ast}$-measurable sets can be strictly larger than the one on which $\mu$ is defined. However, the $\mu$-measurable sets are the same as the $\mu^{\ast}$-measurable sets *provided that* $\mu$ is

*complete* (obviously necessary) and
*$\sigma$-finite* (not strictly necessary but sufficient and good enough for many purposes)

and thus $\mu$ can be obtained back from $\mu^{\ast}$ by Carathéodory's construction (or simply restriction, if you prefer).

The $\sigma$-finiteness condition can be weakened to localizability/local determination conditions, but I prefer to formulate the result for $\sigma$-finite measures, as this is a more widely known condition.

Start with a measure space $(X,\Sigma,\mu)$, complete or not.

It seems to me that in this generality, with no more information at hand, the only thing we can do in the first place is to try and associate an outer measure $\mu^{\ast}$ with that situation. The only one that comes to mind is:

For $A \subset X$ put $\mu^{\ast} (A) = \inf{\left\{\mu(E)\,:\,A \subset E, \,E\in \Sigma\right\}}.$

The following points are all very easy but we'll need them later on, so for the record:

This $\mu^{\ast}$ is an outer measure:

- $\mu^{\ast}(\emptyset) = 0$ and
monotonicity $\mu^{\ast}(A) \leq \mu^{\ast}(B)$ for $A \subset B$ are clear from the definition
- $\sigma$-subadditivity follows from the usual $2^{-n}$-trick:

Let $A = \bigcup_{n=1}^{\infty} A_n$ and $\varepsilon \gt 0$. We may assume that $\mu^{\ast}(A_n)$ is finite for all $n$. For each $A_n$ choose $E_n \in \Sigma$ with $A_n \subset E_n$ and $\mu(E_n) \leq \mu^{\ast}(A_n) + \varepsilon \cdot 2^{-n}$ then $A \subset E = \bigcup_{n=1}^{\infty} E_n$ and $\mu(E) \leq \sum \mu(E_n) \leq \sum \mu^{\ast}(A_n) + \varepsilon$. Since $\varepsilon \gt 0$ was arbitrary, we conclude. $\mu^{\ast}(A) \leq \sum \mu^{\ast}(A_n)$.

For all $E \in \Sigma$ we have $\mu(E) = \mu^{\ast}(E)$.

Notice that the infimum in the definition of $\mu^{\ast}$ is actually a minimum: choose $E_n \in \Sigma$, $A \subset E_n$ such that $\mu(E_n) \leq \mu^\ast(A) + 1/n$. Then $E = \bigcap_{n=1}^{\infty} E_n \in \Sigma$, we have $A \subset E$ and $\mu^{\ast}(A) = \mu(E)$.

The usual Carathéodory extension theorem applied to $\Sigma$ and $\mu$ shows that all the sets $E \in \Sigma$ are measured by $\mu^{\ast}$, so
$$\Sigma \subset \Sigma_{C} = \{E \subset X\,:\,\mu^{\ast}(A) = \mu^{\ast}(A \cap E) + \mu^{\ast}(A \smallsetminus E)\text{ for all } A \subset X\}.$$
Appealing to Carathéodory's extension theorem here is clearly overkill: Let $F \in \Sigma$. By subadditivity we have for all $A \subset X$ that $\mu^{\ast}(A) \leq \mu^{\ast}(A \cap F) + \mu^{\ast}(A \smallsetminus F)$ and if $A$ has infinite outer measure then $\mu^{\ast}(A) \geq \mu^{\ast}(A \cap F) + \mu^{\ast}(A \smallsetminus F)$ holds vacuously, so assume that $\mu^{\ast}(A) \lt \infty$. According to the previous point we can choose $E \in \Sigma$, with $A \subset E$ and $\mu^{\ast}(A) = \mu(E)$. Moreover, $A \cap F \subset E \cap F \in \Sigma$ and $A \smallsetminus F \subset E \smallsetminus F \in \Sigma$ so that $\mu^{\ast}(A \cap F) + \mu^{\ast}(A \smallsetminus F) \leq \mu^{\ast}(E \cap F) + \mu^{\ast}(E\smallsetminus F) = \mu(E \cap F) + \mu(E\smallsetminus F) = \mu(E) = \mu^{\ast}(A)$.

So here's what we have so far: given a measure space $(X,\Sigma,\mu)$ we find an outer measure $\mu^{\ast}$ on $X$, and the associated space $(X,\Sigma_{C},\mu_C)$, where $\mu_{C}$ is the *complete* measure obtained from $\mu^{\ast}$ by Carathéodory's construction. We argued that $\Sigma \subset \Sigma_C$ and for $E \in \Sigma$ we have $\mu(E) = \mu_C(E)$, so $\mu_C$ is an extension of $\mu$.

This gives a first partial affirmative answer to Q1: there always exists an outer measure $\mu^{\ast}$ such that the restriction of $\mu^{\ast}$ to $\Sigma$ is $\mu$ itself. That is, $\mu_{C}$ satisfies requirement 2. of your Q1.

Recall that every measure space $(X,\Sigma,\mu)$ has a *completion*: there is a smallest $\sigma$-algebra $\check{\Sigma}$ and a *complete* measure $\check{\mu}$ such that $\Sigma \subset \check{\Sigma}$ and $\mu(E) = \check{\mu}(E)$ for $E \in \Sigma$.

Explicitly, $\check{\Sigma}$ consists of the sets of the form $E \cup N$ where $E \in \Sigma$ and $N$ is a subset of a $\mu$-null set in $\Sigma$. It is not hard to show that $\check{\Sigma}$ is a $\sigma$-algebra and that every complete measure $\bar{\mu}$ extending $\mu$ must extend $\check{\mu}$. More precisely, if $\bar{\mu}$ is a complete measure defined on a $\sigma$-algebra $\bar{\Sigma} \supset \Sigma$ and $\bar{\mu}(E) = \mu(E)$ for all $E \in \Sigma$ then $\bar{\Sigma} \supset \check{\Sigma}$ and $\bar{\mu}(\check{E}) = \check{\mu}(\check{E})$ for all $\check{E} \in \check{\Sigma}$.

Observe: If $(X,\Sigma,\mu)$ is already complete then $\check{\Sigma} = \Sigma$ and $\check{\mu} = \mu$. If $\mu$ isn't complete, then clearly $\check{\Sigma} \supsetneqq \Sigma$ and $\check{\mu} \neq \mu$ as you already observed by assuming $(X,\Sigma,\mu)$ to be complete.

Finally, it is not very hard to check that $(\check{\mu})^{\ast} = \mu^{\ast}$ so that passing to the completion doesn't change the associated outer measure.

Summarizing what we know so far, we have three measure spaces:

the original measure space $(X,\Sigma,\mu)$ we started with,

its completion $(X,\check{\Sigma},\check{\mu})$,

its extension $(X,\Sigma_{C},\mu_{C})$ obtained via Carathéodory's construction from the outer measure $\mu^{\ast}$.

We know that $\Sigma \subset \check{\Sigma} \subset \Sigma_{C}$ and that $\check{\mu}$ is an extension of $\mu$ and that $\mu_{C}$ is an extension of $\check{\mu}$.

Your question becomes, in my interpretation above:

Are $\check{\Sigma} = \Sigma_C$ and $\check{\mu} = \mu_{C}$?

This turns out to be **wrong** in general.

Here's an artificial, very degenerate, but easily tractable example (I learned this from Fremlin's treatise *Measure theory*, volume 2, exercise 213Ya, page 31):

Let $X$ be a countable set and consider the outer measure $\varphi(A) = \sqrt{\# A}$.

The $\sigma$-algebra of $\varphi$-measurable sets is $\Sigma = \{\emptyset,X\}$: If $\emptyset \neq E \neq X$ choose $A = \{e, x\}$ with $e \in E$ and $x \in X \smallsetminus E$. Then $2 = \varphi(A \cap E) + \varphi(A \smallsetminus E) \gt \varphi(A) = \sqrt{2}$ so that $E$ isn't $\varphi$-measurable. The thing “going wrong here” is of course that $\varphi$ is strictly subadditive on sets of finite non-zero measure.

Clearly $\mu(\emptyset) = 0$ and $\mu(X) = \infty$ is the measure obtained from $\varphi$ by the Carathéodory construction.

Now $\mu^{\ast}$ is the infinite (outer) measure $\mu^{\ast}(\emptyset) = 0$ and $\mu^{\ast}(A) = \infty$ if $A \neq \emptyset$. The $\sigma$-algebra of $\mu^{\ast}$-measurable sets is $\mathcal{P}(X)$ and $\mu_{C} = \mu^{\ast}$.

In particular $\Sigma \subsetneqq \Sigma_{C}$ and $\mu \neq \mu_{C}$.

The take-away is of course:

If $\mu$ is induced from *some* outer measure $\varphi$ then it need not be identical to the measure $\mu_{C}$ given by its associated outer measure $\mu^{\ast}$.

The counterexample above is admittedly very artificial and not something we really care about.

So, we're looking for conditions under which $\check{\mu} = \mu_C$. Here's a natural one, which gives a affirmative answer to Q1 under the *additional* hypothesis that $(X,\Sigma,\mu)$ is $\sigma$-finite, that is to say $X = \bigcup_{n=1}^{\infty} E_{n}$ with $\mu(E_n) \lt \infty$.

The following proposition is a minor adaptation of Fremlin's more general proposition 213C on p.24 of volume 2 of his *Measure theory*:

**Proposition.** If $(X,\Sigma,\mu)$ is *complete* and $\sigma$-finite then $\mu = \mu_{C}$.

In other words, assuming completeness and $\sigma$-finiteness, $\mu$ is induced by the outer measure $\mu^{\ast}$ given by $$\mu^{\ast} (A) = \inf{\left\{\mu(E)\,:\,A \subset E, \,E\in \Sigma\right\}}$$ and $\Sigma$ coincides with the $\mu^{\ast}$-measurable sets in the sense of Carathéodory.

It suffices to prove that $\Sigma = \Sigma_C$. Recall that $\Sigma \subset \Sigma_C$ where $$\Sigma_C = \{E \subset X\,:\,\mu^{\ast}(A) = \mu^{\ast}(A \cap E) + \mu^{\ast}(A \smallsetminus E)\text{ for all } A \subset X\}.$$
Let $F \in \Sigma_C$. Since we assume that $X = \bigcup_{n =1}^{\infty} E_n$ with $E_n \in \Sigma$ and $\mu(E_n) \lt \infty$ we can put $F_n = E_n \cap F$ and it suffices to show that $F_n \in \Sigma$ because $F = \bigcup_{n=1}^{\infty} F_n$.

Choose $G_1 \in \Sigma$ with $F_n \subset G_1 \subset E_n$ and $\mu(G_1) = \mu^{\ast}(E_n \cap F_n)$ and $G_2 \in \Sigma$ with $E_n \smallsetminus F_n \subset G_2 \subset E_n$ and $\mu(G_2) = \mu^{\ast}(E_n \smallsetminus F_n)$. Since $F_n \in \Sigma_C$ we have by definition of $\Sigma_C$ that
$$\mu(G_1) + \mu(G_2) = \mu^{\ast}(E_n \cap F_n) + \mu^{\ast}(E_n \smallsetminus F_n) = \mu^{\ast}(E_n) = \mu(E_n).$$
Since $\mu(E_n) \lt \infty$ and $E_n = G_1 \cup G_2$ we have that $\mu(G_1 \cap G_2) = \mu(G_1) + \mu(G_2) - \mu(E_n) = 0.$
But notice that $G_1 \smallsetminus F_n \subset G_1 \cap G_2$, so $G_1 \smallsetminus F_n$ is a null-set, hence $\mu$-measurable since $\mu$ is complete, and $F_n = G_1 \smallsetminus (G_1 \smallsetminus F_n)$ shows that $F_n \in \Sigma$, as desired.

You showed that $l^*(E) = \inf l^*(\mathcal{O})$ (the infimum runs over all open sets $\mathcal{O}$ containing $E$).

Suppose $E$ is of finite exterior measure -then for any $n\in \mathbb{N}$ we have an open set $\mathcal{O_n}$ such that
$$l^*(\mathcal{O_n}) \leq l^*(E)+n^{-1}<\infty$$
And since $l^*(\mathcal{O_n}) = l^*(E)+l^*(\mathcal{O_n} \setminus E)$ we have $l^*(\mathcal{O_n} \setminus E) \leq n^{-1}$ (The equality is infact a restatement of the measurability of $E$, see Equivalent Definition of Measurable set)

Taking $\mathcal{O} = \bigcap_{n=1}^\infty \mathcal{O_n}$ we have come up with a $G_\delta$ set (countable intersection of open sets) which is obviously a borel set, with the property: $$l^*(\mathcal{O}\setminus E) \leq l^* (\mathcal{O_n}\setminus E) \leq n^{-1}$$ for all $n\in\mathbb{N}$ which is the first corollary.

If $E$ is of infinte exterior measure then we denote $E_n = E \cap B(0,n)$ ($E$'s intersection with the ball of radius $n$ centered at the origin). Each $E_n$ is bounded and so has finite exterior measure (Since it obviously is encompassed within $B(0,n)$) And we may extract $G_n$ sets such that $$l^*(G_n \setminus E_n)=0$$ Putting $G=\bigcup_{n=1}^\infty G_n$ we have $$l^*(G\setminus E) = l^*\left(\bigcup_{n=1}^\infty G_n \setminus \bigcup_{n=1}^\infty E_n \right)\leq \sum_{n=1}^\infty l^*(G_n \setminus E_n ) =0$$

This shows the first corollary.

For the second: For every measurable $E^C$ we obviously have a borel set $G\supset E^C$ such that $l^*(G\setminus E^C) =0$ (Take the set from the first claim) that shows that $G\setminus E^C$ is a lebesgue measurable set as it is a null set.
Then obviously $E^C= G \setminus (G\setminus E^C)= G \cap (G\cap E)^C$. Taking completements on both sides we have:$$ E = G^C \cup (G\setminus E^C)$$

$G^C$ is again a borel set (As a completement of borel set).

As a final comment on this excercise: One notices we used 2 main properties of the Lebesgue measure:

- It is outer regular meaning that for every measurable subset $E$ , for every $\varepsilon>0$ we have an open set $\mathcal{O}\supset E$ with the property $l^*(\mathcal{O} \setminus E) \leq \varepsilon$ which we had to prove.
- It is defined on borel sets, bounded sets have finite exterior-measure.

It turns out that the second implies the first and the first property is what we actually needed. In this proof I made a slight detour (as i wasn't sure which Lebesgue measurability criterion you are using).

## Best Answer

It is by definition. If $B$ satisfies the Carathéodory criterion $\mu^*(T)=\mu^*(T\cap B)+\mu^*(T\cap B^c)$ for every $T\subset\mathbb{R}$, then $B$ is Lebesgue measurable and we may

define$\mu(B)=\mu^*(B)$. Yes, you can call $B$ $\mu^*$-measurable, but this just refers to the outer measure used in the Carathéodory criterion (in case, for example, you are working with more than one outer measure). In general, you just say $B$ is (Lebesgue) measurable.Again, if by this you mean $B$ satisfies the Carathéodory criterion, then yes. If by this question you mean a set $B\subset\mathbb{R}$ such that $\mu^*(B)$ makes sense, then no. Remember, outer measure is defined for all sets in $\mathbb{R}$, and there are certainly sets for which the Carathéodory criterion does not hold.