Let $X$ a set, $\mathcal{A}$ an algebra of subsets in $X$, and $\mu $ a measure in $\mathcal{A}.$
If $B\subset X$, we define $\mu'(B)=\inf\{\mu(A):B\subset A\in\mathcal{A}\}$. I already showed that $\mu'(E)=\mu(E)$ for all $E\in\mathcal{A}$, and now I want to prove that, for all $B\subset X, \mu^{*}(B)\leq\mu'(B),$ where $\mu^{*}(B)=\inf\sum_{n=1}^{\infty}\mu(E_{j}),$ where the infimum is taken by every sequence $(E_{j})\in\mathcal{A}$ such that $B\subset\bigcup_{j=1}^{m}E_{j}.$ Moreover, if $X$ is a countable union of sets with finite $\mu-$measure, $\mu^{*}(B)=\mu'(B).$
My problem is that I can't really see the diference between $\mu^{*}$ and $\mu'$.
(i) If $(E_{j})$ is a sequence of sets in $\mathcal{A}$ with $B\subset\bigcup_{j=1}^{\infty}E_{j},$ I need to show that exists a set $A\in\mathcal{A}$ such that $B\subset A$ and $\mu(A)\leq \sum_{j=1}^{\infty}\mu(E_{j})$. If this is true, I can prove that $\mu^{*}(B)\leq\mu'(B)$.
(ii) On another hand, if $A\in\mathcal{A}$ is a set such that $B\subset A$, so we can define the sequence $(E_{j})\in\mathcal{A}$ by $E_{1}=A$ and $E_{j}=\emptyset$ for all $j>1.$ So $\mu(A)\leq \sum_{j=1}^{\infty}\mu(E_{j}).$ So, I can prove that $\mu'(B)\leq\mu^{*}(B)$.
So, I conclude that $\mu'(B)=\mu^{*}(B).$
How can I prove (i) ? And what was my mistake in (ii)?
To prove $\mu^*(B) \leq \mu'(B),$ we suppose $A \in \mathcal{A}$ such that $B \subseteq A$ and then define $E_1 = A$ and $E_i = \emptyset$ for any $i \geq 2,$ where we note $E_i \in \mathcal{A}$ for every $i$ and $B \subseteq A = \bigcup_{i=1}^\infty E_i.$ This allows us to write $$\mu^*(B) \leq \sum_{i=1}^\infty \mu(E_i) = \mu(A)$$ Since $A \in \mathcal{A}$ such that $B \subseteq A$ was arbitrary, we can now take the infimum to show $$\mu^*(B) \leq \inf \{\mu(A) : A \in \mathcal{A}, B \subseteq A\} = \mu'(B)$$
As I stated in the comments, the reason why we cannot use the above logic to show $\mu'(B) \leq \mu^*(B)$ is that the choice of $E_i \in \mathcal{A}$ was not arbitrarily chosen such that $B \subseteq \bigcup E_i$ but rather it was a very particular choice dependent on the original $A.$