From Rudin's Principles of Mathematical Analysis, we define the outer measure as so:
Definition $11.7$: Let $\mu$ be additive, regular, nonnegative, and finite on $\mathcal{E}$. Consider countable coverings of any set $E \subset \mathbb{R}^n$ by open elementary sets $A_n$:
$$E \subseteq \bigcup_{n=1}^\infty A_n.$$
Define
$$\mu^\ast(E) = \inf \sum_{n=1}^\infty \mu(A_n),$$
the infimum being taken over all countable coverings of $E$ by open elementary sets. $\mu^\ast(E)$ is called the outer measure of $E$, corresponding to $\mu$.
I apologize beforehand as I know there are a multitude of questions regarding the outer-measure, but I have a specific issue that makes the definition a little unclear to me and I hope someone can help me understand it better.
I posted the definition from baby Rudin above for reference.
Now my issue is this, it appears we did not put any restrictions on the coverings of $E$. I am not talking about openness or countability. Rather, the way I am reading the definition is this:
As long as $E \subset \bigcup_{n=1}^{\infty} A_n$, then its outer measure is $$\mu^{*}(E) = \inf \left\{\sum_{n=1}^{\infty} \mu(A_n) \right\} $$
So basically, you can give me a finite cover for a finite compact set $E$, but then add $\mathbb{R}$ to the union of its covers $\bigcup_{n=1}^{N} A_n$, $(N < \infty)$, which is true, then
$$\mu^{*}(E) = \inf \left\{\sum_{n=1}^{N} \mu(A_n) + \mu\mathbb{R} \right\} = inf\{\infty \} = \infty$$
Best Answer
You're missing a key specification in the definition of outer measure:
The covering $\{A_n\}_{n=1}^\infty$ of $E$ must be by elementary (open) sets. Elementary sets, as noted in Definition $11.4$, are those which are intervals or finite unions of intervals (with intervals being the Cartesian product of closed, open, or half-open intervals in the usual sense in $\mathbb{R}^1$).
Now see that $\mathbb{R}$ is not an elementary set, so you can't just include it in a covering. To wit, suppose otherwise, that $\newcommand{\R}{\mathbb{R}} \R$ is the finite union of intervals, and we may limit our consideration to intervals in $\R$ for obvious reasons.
Thus, $\R$ is elementary iff, without loss of generality,
$$\R = \bigcup_{i=1}^N [a_i,b_i]$$
for some $\newcommand{\N}{\mathbb{N}} N \in \N$ for some $a_i,b_i \in \R$ with $a_i \le b_i$.
But clearly, the point
$$\alpha := -1 + \min_{1 \le i \le N} a_i$$
is not in this union, so $\R$ can't be equal to that. Hence $\R$ is not elementary.
Also, another key element overlooked is that the infimum is over all possible countable covers by open, elementary sets. All of them.
Given $E \newcommand{\CC}{\mathcal{C}} \newcommand{\AA}{\mathcal{A}}$, define
$$\CC_E := \left\{ \AA := \left\{ A_n \right\}_{n=1}^\infty \, \middle| \, A_n \text{ is open & elementary $\forall n \in \N$, and $\displaystyle \bigcup_{n=1}^\infty A_n = E$} \right\}$$
That is, $\CC_E$ is the class of all countable coverings of $E$ which use open, elementary sets.
Then the outer measure is defined by
$$\mu^\ast(E) := \inf_{\AA \in \CC_E} \sum_{i=1}^\infty m(A_i)$$
where $m(A_i)$ is defined as in Rudin's text: if $A$ is an elementary set and a finite disjoint union of intervals $I_i$, then $m(A) = \sum m(I_i)$.
With these formalities out of the way, think about what this means -- and in particular, the implications of using infimum. We need the smallest possible value (if just in a limiting sense), of $\sum m(A_i)$ for some cover. Even if $\R$ were elementary, that doesn't matter if you can cover $E$ by some other sets and get a $\sum m(A_i)$ of finite value.
Different values may be generated by different covers, and $\mu^\ast(E)$ wants the smallest of them all.
After all, we would say that
$$\inf (0,1) = 0 \quad \text{ i.e. } \quad \inf_{x \in (0,1)} x = 0$$
in spite of the fact that $1/2$ is in the interval. Sure, it's in the interval, but it's bigger than a lot of other elements in $(0,1)$, so it can't be the infimum.
As an explicit example, we could cover $\N$ by the elementary sets (and disjoint intervals) $B_{n,\varepsilon} := (n - \varepsilon/2^n,n+\varepsilon/2^n)$ for $\varepsilon \in (0,1/2)$. This would form a countable cover of $\N$ by disjoint, open, elementary sets: in fact, each fixed $\varepsilon$ generates such a cover. And notice, for that specific class of covers depending on $\varepsilon$
$$m \left( B_{n,\varepsilon} \right) = \frac{\varepsilon}{2^{n-1}}$$
so
$$\sum_{n=1}^\infty m \left( B_{n,\varepsilon} \right) = \sum_{n=1}^\infty \frac{\varepsilon}{2^{n-1}} = 2\varepsilon$$
using the geometric series.
And again, each $\varepsilon$ is tied to a different covering of $\N$. Clearly, based on the above, smaller $\varepsilon$'s will generate covers with smaller cumulative size.
It does not matter that the $\varepsilon = 1/4$ case gives a cover of total size $1/2$, when the cover corresponding to $\varepsilon = 1/100$ gives one of total size $1/50$, which is far smaller.
In fact, definitionally,
$$\inf_{\AA \in \CC_E} \sum_{A_i \in \AA} m(A_i) \le \sum_{B_i \in \mathcal{B}} m(B_i)$$
for any particular cover $\mathcal{B} := \{B_i\}_{n=1}^\infty$ in $\CC_E$: this is just a consequence the definition of infimum! (After all, we can't rule out the possibility of some other covering having a smaller total size.) In fact, that means that, in the analogy for $\N$ above,
$$\mu^\ast(\N) \le 2 \varepsilon$$
for any particular $\varepsilon \in (0,1/2)$. Of course, since the outer measure is definitionally nonnegative, you can see what happens if we take $\varepsilon$ arbitrarily small: we get $\mu^\ast(\N) = 0$.
This also shows what happens if we add set of incredibly large size to the cover. $\R$ is not elementary, but I can add in the interval $I := [-10^{100},10^{100}]$ with no issue, i.e. the cover $\newcommand{\ve}{\varepsilon} \mathcal{B} := \{ B_{n,\ve}\}_{n=1}^\infty$ now becomes $\mathcal{D} := \{ B_{n,\ve}\}_{n=1}^\infty \cup \{I\}$. This is still a totally valid covering: we have $\mathcal{D} \in \CC_\N$. But clearly,
$$2 \varepsilon = \sum_{B_{n,\ve} \in \mathcal{B}} m(B_{n,\ve}) \le \sum_{D_i \in \mathcal{D}} m(D_i) \le m(I) + \sum_{B_{n,\ve} \in \mathcal{B}} m(B_{n,\ve}) = 2 \cdot 10^{100} + 2 \varepsilon$$
so while we added $I$ to the covering, it generated a much larger value. So we don't really care -- because we have a smaller value of just $2 \ve$, that's the one of more concern when it comes to the infimum. Between the two covers $\mathcal{B,D}$, only $\mathcal{B}$ has a chance to be the infimum, and hence the outer measure of $\mu^\ast(\N)$.
In summary: