Question on the definition of the outer measure For a set $\Omega$, let $\mathcal A$ be the set that contains every set $A \in P(\Omega)$ with $A_1$ $\subset A_2$ $\subset$ $ ... $ $~$ and $\bigcup_{k=1}^\infty A_k = A$, $A_k \in \mathscr A.$ ($\sigma$-algebra)
(Side note: in class, we wrote that with an arrow showing upwards, but I didn't found anything for it in the tutorial. Is there a manual for it though?) 
Furthermore, let $\phi: \mathscr A \rightarrow \Bbb R_{+}$ be a measure that we continue on $\mathcal A$ and note it as $\phi^{*}$, the outer measure. For every $X \subset \Omega$, we define it as
$\phi^{*}(X)$ $:=$ $\inf\{\phi(A) : X \subset A, A \in \mathcal A \}.$
My question is the following: When I understood it correctly, the outer measure is the measure of some $A \in \mathcal A$ such that $X$ is still a subset of that specific $A$. If we'd visualise this with circles, we would try to make the circle $A$ as small as possible while it still has to be a super set of $X$. But I wonder, since this is noted as an infinum, if there can't be a case where we could simply choose $A$ such that $X = A$. For me, it looks like we can always only "approximate" $X$ with the help of $A$ by "shrinking" it.
I guess both is true, isn't it? There might be cases where we could choose $X = A$, but most of the time, we can only approximate X. But why is that? I find it hard to work on those different tiers. How are $X$ and $A$ actually connected? 
 A: I find your post hard to read, but I think you are getting to the main idea of the outer measure.
The main point is that the measure $\phi$ is defined only on a subset $\mathscr A$ of the power set on $\Omega$. $\mathscr A$ is typically a $\sigma$-algebra, but can also be simply an algebra (field) or some other subset.
In contrast, the outer measure $\phi^*$ is defined for any set $A \subset \Omega$. Hence you can always write $\phi^*(A)$ for any set $A$ and this is always well defined.
On the other hand, you can only write $\phi(A)$ for sets $A \in \mathscr{A}$, it is not even defined for other sets.
One of the keys properties one proves immediately after the definitions is that (when $\mathscr{A}$ is an algebra, as LeGrandDODOM pointed out):
$$\phi^*(A) = \phi(A) \text{ for }  A \in \mathscr{A}$$
Hence for $A \in \mathscr{A}$ (and $\mathscr{A}$ an algebra), you are correct, to get the outer measure you cover $A$ with itself and this gives the tightest result. But again, for $A$ not in $\mathscr{A}$ you have to cover it with sets which are actually in $\mathscr{A}$ and you cannot cover it with itself.
