Can we say that $\mu^* (A) = \mu (A),$ for all $A \in \mathcal A$? Let $\mathcal A$ be an algebra of subsets of a set $X$ and $\mu : \mathcal A \longrightarrow [0,+\infty]$ be a measure on $\mathcal A.$ Define $\mu^* : \mathcal P(X) \longrightarrow [0,+\infty]$ by $$\mu^* (A) : = \text {inf}\ \left \{ \sum\limits_{j=1}^{\infty} \mu \left (A_j \right )\ :\ \bigcup\limits_{j=1}^{\infty} A_j \supseteq A,\ A_j \in \mathcal A \right \},\ \ \ \ A \subseteq X.$$ Then clearly $\mu^*$ is well-defined and it is called the outer measure of subsets of $X.$
With this definition in mind we can easily come up with some of the properties of $\mu^*.$
$(1)$ $\mu^* (\varnothing) = 0$ and $\mu^*(A) \geq 0,$ for all $A \subseteq X.$
$(2)$ $\mu^*$ is monotonic i.e. if $A \subseteq B \subseteq X$ then $\mu^* (A) \leq \mu^* (B).$
$(3)$ $\mu^*$ is countably sub-additive i.e. given $A \subseteq X$ and given a countable covering $\left \{A_j \right \}_{j=1}^{\infty}$ of $A$ by subsets of $X$ we have $$\mu^* (A) \leq \sum\limits_{j=1}^{\infty}\mu^*(A_j).$$
Also it is clear that $\mu^* (A) \leq \mu(A)$ as $\{A \}$ itself a countable covering of $A$ by elements of $\mathcal A.$ Can we say the other part of the inclusion to be true? i.e. can I say that $\mu(A) \leq \mu^*(A)$? I have watched a video recently in youtube which proves the result to be true. But I found a mistake in it. It argues as follows $:$
If $\mu^* (A) = +\infty$ then we have nothing to prove. So WLOG let us assume that $\mu^*(A) < +\infty.$ Let us fix $\varepsilon > 0$ arbitrarily. Then by the property of infimum $\exists$ a countable covering $\{A_j \}_{j=1}^{\infty}$ of $A$ by elements of $\mathcal A$ such that $$\mu^*(A) + \varepsilon > \sum\limits_{j=1}^{\infty} \mu (A_j).$$
Making $\varepsilon \to 0$ we have $$\mu^*(A) \geq \sum\limits_{j=1}^{\infty} \mu (A_j).$$
Now $\mu$ being a measure on $\mathcal A$ it is countably sub-additive. So we have $$\mu (A) \leq \sum\limits_{j=1}^{\infty} \mu (A_j).$$
This is the stage where I didn't understand the reasoning. How do we say that $\bigcup\limits_{j=1}^{\infty} A_j \in \mathcal A$? $\mathcal A$ being an algebra it is only closed under finite unions but might not be closed under countable unions. Right? But then we cannot certainly invoke the concept of countable sub-additivity of $\mu$ here. Can anybody please make it clear to me? Any helpful suggestion will be highly appreciated.
Thank you very much for your valuable time for reading.
Source $:$ https://youtu.be/6_RQ1c1VVBs?list=PLtKWB-wrvn4mbGE2XeUbnVw1cwAj4-f0C&t=699
 A: Indeed, we cannot assert that $\bigcup_{j=1}^\infty A_j \in \mathcal{A}$.  But we don't have to.
First note that $\mu$ enjoys the following countable sub-additivity property:

(*) If $B_j, B \in \mathcal{A}$ and $B = \bigcup_{j=1}^\infty B_j$, then $\mu(B) \le \sum_{j=1}^\infty \mu(B_j)$.

This is proved by the usual "disjointification": define $C_j$ inductively as $C_1 = B_1$, $C_{j+1} = B_{j+1} \setminus \bigcup_{i=1}^j C_i$.  Then the $C_j$ are disjoint and $B = \bigcup_{j=1}^\infty C_j$ so by countable additivity $\mu(B) = \sum_{j=1}^\infty \mu(C_j)$.  But we have $C_j \subseteq B_j$ so by finite additivity $\mu(C_j) \le \mu(B_j)$, hence $ \sum_{j=1}^\infty \mu(C_j) \le \sum_{j=1}^\infty \mu(B_j)$.
Now, returning to the original problem, apply property (*) with $B_j = A \cap A_j$.
We get $\mu(A) \le \sum_{j=1}^\infty \mu(B_j)$, and since $B_j \subseteq A_j$, we conclude $\mu(A) \le \sum_{j=1}^\infty \mu(A_j)$ as desired.
So we do indeed have the following improved countable sub-additivity property:

(**) If $A_j, A \in \mathcal{A}$ and $A \subseteq \bigcup_{j=1}^\infty A_j$, then $\mu(A) \le \sum_{j=1}^\infty \mu(A_j)$.  This holds whether $\bigcup_{j=1}^\infty A_j \in \mathcal{A}$ or not.

This justifies the line of the proof that you were asking about.
