Concerning the first question, it can be done for the Lebesgue measure, but the proof is long. I taught it like that once, but I was not happy with it.
I am not taking the intervals to be open (it is equivalent).
$$
\mu^{\ast}(E):=\inf\left\{ \sum_{i=1}^{\infty}\ell(R_{i}):~R_{i}\text{
intervals, }\bigcup_{i=1}^{\infty}R_{i}\supseteq E\right\} .
$$
Given a set $E\subseteq\mathbb{R}$, we say that $E$ is Lebesgue
measurable if for every $\varepsilon>0$ there exists an open set $U\supseteq
E$ such that
$$
\mu^{\ast}(U\setminus E)\leq\varepsilon.
$$
Proposition 1
The following properties hold.
(i) If $E\subseteq F\subseteq\mathbb{R}$, then $\mu^{\ast}(E)\leq
\mu^{\ast}(F)$.
(ii) If $E\subseteq\bigcup_{n=1}^{\infty}E_{n}$, then $\mu^{\ast
}(E)\leq\sum_{n=1}^{\infty}\mu^{\ast}(E_{n})$.
(iii) If $E\subseteq\mathbb{R}$, then
$$
\mu^{\ast}(E)=\inf\{\mu^{\ast}(U):~U\text{ open, }U\supseteq E\}.
$$
(iv) If $E,F\subseteq\mathbb{R}$ and $\operatorname*{dist}(E,F)>0$,
then $\mu^{\ast}(E\cup F)=\mu^{\ast}(E)+\mu^{\ast}(F)$.
Proof: (i) If $R_{i}$ are intervals such that $\bigcup_{i=1}^{\infty}
R_{i}\supseteq F$, then $\bigcup_{i=1}^{\infty}R_{i}\supseteq E$, and so
$$
\mu^{\ast}(E)\leq\sum_{i=1}^{\infty}\ell(R_{i}).
$$
Taking the infimum over all sequences of intervals covering $F$ gives
$\mu^{\ast}(E)\leq\mu^{\ast}(F)$.
(ii) If $\sum_{n=1}^{\infty}\mu^{\ast}(E_{n})=\infty$, there is nothing to
prove. Thus, assume that $\sum_{n=1}^{\infty}\mu^{\ast}(E_{n})<\infty$. Given
$\varepsilon>0$, for each $n$ find a sequence $\{R_{i}^{\left( n\right)
}\}_{i}$ of intervals such that $\bigcup_{i=1}^{\infty}R_{i}^{(n)}\supseteq
E_{n}$ and such that
$$
\sum\limits_{i=1}^{\infty}\ell(R_{i}^{(n)})\leq\mu^{\ast}(E_{n})+\frac
{\varepsilon}{2^{n}}.
$$
Since $\mathbb{N}\times\mathbb{N}$ is countable, we may write $\{R_{i}
^{(n)}\}_{i,n\in\mathbb{N}}=\left\{ R_{j}\right\} _{j\in\mathbb{N}}$. Note
that
$$
\bigcup_{n=1}^{\infty}E_{n}\subset\bigcup_{j=1}^{\infty}R_{j}=\bigcup
\limits_{n=1}^{\infty}\bigcup\limits_{i=1}^{\infty}R_{i}^{\left( n\right) },
$$
and so by part (i) and (by properties of nonnegative double series)
$$
\mu^{\ast}(E)\leq\mu^{\ast}\left( \bigcup\limits_{n=1}^{\infty}E_{n}\right)
\leq\sum\limits_{j=1}^{\infty}\ell(R_{j})=\sum\limits_{n=1}^{\infty}
\sum\limits_{i=1}^{\infty}\ell(R_{i}^{\left( n\right) })\leq\sum
\limits_{n=1}^{\infty}\mu^{\ast}(E_{n})+\varepsilon.
$$
By letting $\varepsilon\rightarrow0^{+}$ we conclude the proof.
(iii) If $U$ is open and $U\supseteq E$, then by part (i), $\mu^{\ast}
(E)\leq\mu^{\ast}\left( U\right) $. Taking the infimum over all open sets
containing $E$ gives
$$
\mu^{\ast}(E)\leq\inf\left\{ \mu^{\ast}(U):~U\text{ open, }U\supseteq
E\right\} .
$$
On the other hand, if $R_{i}$ are intervals such that $\bigcup_{i=1}^{\infty
}R_{i}\supseteq E$, given $\varepsilon>0$ for every $i$ consider an open
interval $T_{i}\supseteq R_{i}$ such that $\operatorname*{meas}T_{i}
\leq\operatorname*{meas}R_{i}+\frac{\varepsilon}{2^{i}}$ and let
$V:=\bigcup_{i=1}^{\infty}T_{i}$. Then $V$ is open and contains $E$ and so
$$
\inf\left\{ \mu^{\ast}(U):~U\text{ open, }U\supseteq E\right\} \leq\mu
^{\ast}(V)\leq\sum_{i=1}^{\infty}\ell(T_{i})\leq\sum_{i=1}^{\infty}\ell
(R_{i})+\varepsilon.
$$
Letting $\varepsilon\rightarrow0^{+}$ gives
$$
\inf\left\{ \mu^{\ast}(U):~U\text{ open, }U\supseteq E\right\} \leq
\sum_{i=1}^{\infty}\ell(R_{i}).
$$
Taking the infimum over all sequences of intervals covering $E$ gives
$$
\inf\left\{ \mu^{\ast}(U):~U\text{ open, }U\supseteq E\right\} \leq\mu
^{\ast}(E).
$$
(iv) Let $d:=\operatorname*{dist}(E,F)$. If $R_{i}$ are intervals such that
$\bigcup_{i=1}^{\infty}R_{i}\supseteq E\cup F$, by subdividing each interval
$R_{i}$ into subintervals, without loss of generality, we may assume that each
interval $R_{i}$ has diameter less than $d$. Hence, if $R_{i}$ intersects $E$
it cannot intersect $F$ and viceversa. Write $\{R_{i}:~R_{i}\cap
E\neq\emptyset\}=\{S_{i}\}$ and $\{R_{i}:~R_{i}\cap F\neq\emptyset
\}=\{T_{i}\}$. Then $\bigcup_{i}S_{i}\supseteq E$, $\bigcup_{i}T_{i}\supseteq
F$ and so
$$
\mu^{\ast}(E)+\mu^{\ast}(F)\leq\sum_{i}\ell(S_{i})+\sum_{i}\ell(T_{i}
)=\sum_{i=1}^{\infty}\ell(E_{i}).
$$
Taking the infimum over all sequences of intervals covering $E\cup F$ gives
$$
\mu^{\ast}(E)+\mu^{\ast}(F)\leq\mu^{\ast}(E\cup F).
$$
The other inequality follows from part (ii).
Proposition 2
The following properties hold.
(i) Open sets are Lebesgue measurable.
(ii) If $E\subseteq\mathbb{R}$ has Lebesgue outer measure zero, then
$E$ and its subsets are Lebesgue measurable.
(iii) If $E=\bigcup_{n=1}^{\infty}E_{n}$, and each $E_{n}$ is Lebesgue
measurable, then $E$ is Lebesgue measurable.
(iv) Compact sets are Lebesgue measurable.
(v) Closed sets are Lebesgue measurable.
(vi) If $E\subseteq\mathbb{R}$ is Lebesgue measurable, then
$\mathbb{R}\setminus E$ is Lebesgue measurable.
(vii) If $E=\bigcap_{n=1}^{\infty}E_{n}$, and each $E_{n}$ is Lebesgue
measurable, then $E$ is Lebesgue measurable.
Proof (ii) Let $E\subseteq\mathbb{R}$ be such that $\mu^{\ast}(E)=0$. By
Proposition 1(iii),
$$
0=\mu^{\ast}(E)=\inf\left\{ \mu^{\ast}(U):~U\text{ open, }U\supseteq
E\right\} .
$$
Hence, for every $\varepsilon>0$ there exists $U$ open such that $U\supseteq
E$ and $\mu^{\ast}(U)\leq\varepsilon$. By Proposition
1(i), $\mu^{\ast}(U\setminus E)\leq
\mu^{\ast}(U)\leq\varepsilon$, and thus $E$ is Lebesgue measurable.
Again by Proposition 1(i), if $F\subseteq
E$, then $0\leq\mu^{\ast}(F)\leq\mu^{\ast}(E)=0$, and so $F$ also has Lebesgue
outer measure zero, and thus is Lebesgue measurable.
(iii) Let $E=\bigcup_{n=1}^{\infty}E_{n}$, where each $E_{n}$ is Lebesgue
measurable. Then for every $\varepsilon>0$ there exists an open set
$U_{n}\supseteq E_{n}$ such that
$$
\mu^{\ast}(U_{n}\setminus E_{n})\leq\frac{\varepsilon}{2^{n}}.
$$
Define $U:=\bigcup_{n=1}^{\infty}U_{n}$. Then $U$ is open, contains $E$ and
$U\setminus E=\bigcup_{n=1}^{\infty}(U_{n}\setminus E)\subseteq\bigcup
_{n=1}^{\infty}(U_{n}\setminus E_{n})$. Hence, by Proposition
1(i) and (ii),
$$
\mu^{\ast}(U\setminus E)\leq\sum\limits_{n=1}^{\infty}\mu^{\ast}%
(U_{n}\setminus E_{n})\leq\varepsilon.
$$
(iv) Let $K\subset\mathbb{R}$ be a compact set. Then $K$ is bounded and so
$\mu^{\ast}(K)<\infty$. Then by Proposition
1(iii) for every $\varepsilon>0$ there
exists an open set $U\supseteq K$ such that
$$
\mu^{\ast}(U)\leq\mu^{\ast}(K)+\varepsilon.
$$
The set $U\setminus K$ is open and so we can write
$$
U\setminus K=\bigcup_{n=1}^{\infty}Q_{n},
$$
where $Q_{n}$ are closed bounded intervals with disjoint interior. Hence, for
every $m\in\mathbb{N}$ the set
$$
C_{m}:=\bigcup_{n=1}^{m}Q_{n}
$$
is compact and does not intersect $K$. Hence, $\operatorname*{dist}%
(C_{m},K)>0$. Since $K\cup C_{m}\subseteq U$, by Proposition
1(i), (iv),
\begin{align*}
\mu^{\ast}(U) & \geq\mu^{\ast}(K\cup C_{m})=\mu^{\ast}(K)+\mu^{\ast}(C_{m})\\
& =\mu^{\ast}(K)+\sum_{n=1}^{m}\ell(Q_{n})
\end{align*}
Hence,
$$
\sum_{n=1}^{m}\ell(Q_{n})\leq\mu^{\ast}(U)-\mu^{\ast}(K)\leq\varepsilon
$$
for all $m$. Letting $m\rightarrow\infty$ gives
$$
\mu^{\ast}(U\setminus K)\leq\sum_{n=1}^{\infty}\ell(Q_{n})\leq\mu^{\ast
}(U)-\mu^{\ast}(K)\leq\varepsilon,
$$
which shows that $K$ is measurable.
(v) If $C\subseteq\mathbb{R}$ is closed, then $C=\bigcup_{n=1}^{\infty}
(C\cap\lbrack-n,n])$ and since each $C\cap\lbrack-n,n]$ is compact, it follows
from parts (iii) and (iv) that $C$ is measurable.
(vi) If $E\subseteq\mathbb{R}$ is Lebesgue measurable, then for every
$n\in\mathbb{N}$ there exists an open set $U_{n}\supseteq E$ such that
$$
\mu^{\ast}(U_{n}\setminus E)\leq\frac{1}{n}.
$$
By part (v) the set $C_{n}:=\mathbb{R}\setminus U_{n}$ is closed and so
Lebesgue measurable. Define $F:=\bigcup_{n=1}^{\infty}C_{n}$. Then $F$ is
Lebesgue measurable by part (iii). Since $U_{n}\supseteq E$, we have that
$\mathbb{R}\setminus E\supseteq\mathbb{R}\setminus U_{n}=C_{n}$ for every $n$,
and so $\mathbb{R}\setminus E\supseteq F$. Moreover,
$$
(\mathbb{R}\setminus E)\setminus F=(\mathbb{R}\setminus E)\setminus
\bigcup_{n=1}^{\infty}C_{n}\subseteq U_{n}\setminus E
$$
for every $n$ and so
$$
\mu^{\ast}((\mathbb{R}\setminus E)\setminus F)\leq\mu^{\ast}(U_{n}\setminus
E)\leq\frac{1}{n}.
$$
Letting $n\rightarrow\infty$ shows that $\mu^{\ast}((\mathbb{R}\setminus
E)\setminus F)=0$ and so $(\mathbb{R}\setminus E)\setminus F$ is measurable by
part (ii). Since $F$ is measurable, we have that $\mathbb{R}\setminus E$ is
Lebesgue measurable since it is the union of $F$ and $(\mathbb{R}\setminus
E)\setminus F$.
(vii) This follows from part (vi) and De Morgan's laws.
Remark
In the proof of part (iv) we used the fact that
$$
\mu^{\ast}(C_{m})=\sum_{n=1}^{m}\ell(Q_{n}).
$$
To see this, if $Q_{n}$ has length $\ell_{n}$, consider the intervals with the
same center $S_{n}$ of length $(1-\varepsilon)\ell_{n}$. Since the intervals
$Q_{n}$ have pairwise disjoint interior, it follows that $\operatorname*{dist}
(S_{n},S_{k})>0$ for $n\neq k$ and so by Propositions
1
(iv)
\begin{align*}
\mu^{\ast}(C_{m}) & \geq\mu^{\ast}\Big(\bigcup_{n=1}^{m}S_{n}\Big)=\sum
_{n=1}^{m}\mu^{\ast}(S_{n})\\
& =\sum_{n=1}^{m}\ell(S_{n})=\sum_{n=1}^{m}(1-\varepsilon)^{N}\ell(Q_{n}).
\end{align*}
Letting $\varepsilon\rightarrow0^{+}$ gives
$$
\mu^{\ast}(C_{m})\geq\sum_{n=1}^{m}\ell(Q_{n}).
$$
The other inequality follows from Proposition 1(ii).
Best Answer
Yes there is! Consider the function $\mu^{*}:P(\mathbb{R})\to [0,\infty)$ given by
$$\mu^{*}(A)= \begin{cases} 1, & \text{if} \hspace{2mm} A\neq \emptyset, \\ 0, & \text{if} \hspace{2mm} A=\emptyset . \end{cases}$$ It is easy to see that $\mu^{*}$ is an outer measure on $\mathbb{R}$. Now, I claim that the only two subsets of $\mathbb{R}$ satisfying the Carathéodory criterion are $\mathbb{R}$ and $\emptyset$. Clearly, if we let $E=\mathbb{R}$ or $E=\emptyset$ then, for every $A\subseteq \mathbb{R}$, the following equality is true $$\mu^{*}(A)= \mu^{*}(A\cap E) + \mu^{*}(A\setminus E).$$
Conversely, suppose that $\emptyset\neq E \subset \mathbb{R}$ (here the symbol $\subset$ means proper inclusion). Let $a\in E$, $b\in \mathbb{R}\setminus E$ and put $A=\{a,b\}$. Then, $$\mu^{*}(A)= 1 \neq 2 = 1+1 =\mu^{*}(A\cap E) + \mu^{*}(A\setminus E).$$
As a final comment, if we consider any set $X$ with at least two elements, then the corresponding $\mu^{*}$ produces an outer measure on $X$ such that the only $\mu^{*}$-measurable sets are the empty set and $X$ itself.