In my book it is defined as follows: If E is any subset of $\mathbb{R}^d$, the exterior measure of $E$ is
$$ m_*(E)=\inf\sum _{j=1}^{\infty } |Q_j|$$
where the infimum is taken over all countable coverings by closed cubes($Q$)
Second useful thing: $a \le b$ and $b \le a$, then $a=b$.
The proof where this is used:
My question becomes:
Now this is what I don't get: why is $(1+\epsilon) \sum _{j=1}^{\infty } |Q_j|$ equal to $$\inf\sum _{j=1}^{\infty } |Q_j|$$
In other words I dont get how this is proven:
Second useful thing: method of proof: $a\le b$ and $b \le a$, then $a=b$.
Because when we try to prove $b \le a$ we need to introduce $\epsilon$ but even if this is arbitrary it is still larger than zero.
After being provided with a answer I initially thought I had understood it, but I now have further questions:
If for a fixed $\epsilon$ we choose for each j a cube $Q_j$$\subset$$S_j$ and such that $$\left| S_j\right| \leq (1+e) \left| Q_j\right|$$
Then I have a struggle with the following:
If I am using this to provide my proof then I need to be sure that this condition exist:
My assumption is(please let me know if I am wrong!)
If $Q_j$$\subset$$S_j$, then $\left| Q_j\right|<=\left|S_j\right|$
Now if that assumption is correct:
then choose $a=\left|S_j\right|$
and $b=\left|Q_j\right|$
Now if $$a \leq (1+e) b$$
then $$a \leq b$$
proof: assume the opposite: $$b <a$$ then a-b>0, so we can choose $\epsilon=a-b$
it the follows that $$a \leq b$$ and we end up with a contradiction.
Now this to me means that when they conclude that since $\epsilon$ is arbitrary it cant be because it is fixed and therefore has to be a specific?
It is not the case that $(1 + \epsilon) \sum \left\vert Q_j \right\vert$ is equal to $\inf \sum \left\vert Q_j \right\vert$. (Indices of summation/union ommitted where obvious.)
What is true is that for any $\epsilon > 0$ and any collection $\left\{ Q_j \right\}$ such that $Q \subset \bigcup Q_j$, we must have that
$$ \vert Q \vert \le (1 + \epsilon) \sum \left\vert Q_j \right\vert.$$
This implies (as noted in the comments) that
$$ \vert Q \vert \le \sum \left\vert Q_j \right\vert.$$
Since the collection $\left\{ Q_j \right\}$ was arbitrary (among coverings of $Q$),
$$ \vert Q \vert \le \inf \sum \left\vert Q_j \right\vert.$$
To see that $a \le (1 + \epsilon) b$ for every $\epsilon > 0$ implies that $a \le b$, suppose that $b < a$. Convince yourself that you can find an $\epsilon > 0$ such that $(1 + \epsilon) b < a$.