Prove that $\mu^*(B\cap(A_1\cup A_2)) = \mu^*(B\cap A_1)+\mu^*(B\cap A_2)$. I am reading a Lebesgue integration book.
There is the following proposition without a proof in this book:

Let $\mu^* : 2^\mathbb{R} \to [0,\infty]$ be the Lebesgue outer measure on $\mathbb{R}$.
Let $\mathcal{M} \subset 2^\mathbb{R}$ be the set of all measurable sets.
Let $A_1,\dots,A_n\in\mathcal{M}$.
Assume that $A_i \cap A_j=\emptyset$ for $i\ne j$.
Then, for any $B\in 2^\mathbb{R}$, $$\mu^*(B\cap\bigcup_{i=1}^n A_i) = \sum_{i=1}^n \mu^*(B\cap A_i)$$ holds.

I tried to prove this proposition, but I was not able to prove it.
My attempt is here:
Let $B\in 2^\mathbb{R}$.
Since $A_1\cup A_2\in\mathcal{M}$, $\mu^*(B)=\mu^*(B\cap(A_1\cup A_2)) + \mu^*(B\cap(A_1\cup A_2)^C)$.
Since $A_1\in\mathcal{M}$, $\mu^*(B)=\mu^*(B\cap A_1)+\mu^*(B\cap A_1^C)$.
Since $A_2\in\mathcal{M}$, $\mu^*(B\cap A_1^C)=\mu^*((B\cap A_1^C)\cap A_2)+\mu^*((B\cap A_1^C)\cap A_2^C)$.
So, $\mu^*(B)=\mu^*(B\cap A_1)+\mu^*((B\cap A_1^C)\cap A_2)+\mu^*((B\cap A_1^C)\cap A_2^C)$.
Since $A_1\cap A_2=\emptyset$, $A_2\subset A_1^C$. So, $A_1^C \cap A_2 = A_2$.
So, $\mu^*(B)=\mu^*(B\cap A_1)+\mu^*(B\cap A_2)+\mu^*((B\cap A_1^C)\cap A_2^C)$.
Since $C:=B\cap(A_1\cup A_2)^C=(B\cap A_1^C)\cap A_2^C$, $\mu^*(B\cap(A_1\cup A_2)) + \mu^*(C) = \mu^*(B\cap A_1)+\mu^*(B\cap A_2)+\mu^*(C)$.
If $\mu^*(C) \in\mathbb{R}$, then $\mu^*(B\cap(A_1\cup A_2)) = \mu^*(B\cap A_1)+\mu^*(B\cap A_2)$.
But I cannot prove that $\mu^*(B\cap(A_1\cup A_2)) = \mu^*(B\cap A_1)+\mu^*(B\cap A_2)$ when  $\mu^*(C)=\infty$.
 A: As you proved, when $\mu^*(C)<\infty$ we have $\mu^*(B\cap(A_1\cup A_2)) = \mu^*(B\cap A_1)+\mu^*(B\cap A_2)$.   $(\star)$
Now prove the proposition in two case:

*

*If $\mu^*(B\cap(A_1\cup A_2))=\infty$, since $\mu^*(B\cap A_1)+\mu^*(B\cap A_2)\ge\mu^*(B\cap(A_1\cup A_2))=\infty $, it holds true obviously;


*If $\mu^*(B\cap(A_1\cup A_2))<\infty$, we put $B\cap(A_1\cup A_2)$ in $(\star)$ instead of $B$, it follows that
$$C=(B\cap(A_1\cup A_2))\cap(A_1\cup A_2)^C=\emptyset.$$
A: Your answer is almost there. Here is how to simplify and complete it.
Let $D\in 2^\mathbb{R}$.
Since $A_1\in\mathcal{M}$, $\mu^*(D)=\mu^*(D\cap A_1)+\mu^*(D\cap A_1^C)$.
Since $A_2\in\mathcal{M}$, $\mu^*(D\cap A_1^C)=\mu^*((D\cap A_1^C)\cap A_2)+\mu^*((D\cap A_1^C)\cap A_2^C)$.
So, $\mu^*(D)=\mu^*(D\cap A_1)+\mu^*((D\cap A_1^C)\cap A_2)+\mu^*((D\cap A_1^C)\cap A_2^C)$.
Since $A_1\cap A_2=\emptyset$, $A_2\subset A_1^C$. So, $A_1^C \cap A_2 = A_2$. Also, we have  $(D\cap A_1^C)\cap A_2^C = D\cap (A_1^C\cap A_2^C)  = D\cap (A_1 \cup A_2)^C$.
So, $$\mu^*(D)=\mu^*(D\cap A_1)+\mu^*(D\cap A_2)+\mu^*(D\cap (A_1 \cup A_2)^C)$$
Now, since $D$ is any subset of $\mathbb{R}$, take $D= B\cap(A_1 \cup A_2)$.
Since $B\cap(A_1 \cup A_2) \cap A_i= B\cap A_i$, for $i=1,2$, and $B\cap(A_1 \cup A_2) \cap (A_1 \cup A_2)^C= \emptyset$,  we have
$$\mu^*(B\cap(A_1 \cup A_2))=\mu^*(B\cap A_1)+\mu^*(B\cap A_2)+\mu^*(\emptyset)$$ that is
$$\mu^*(B\cap(A_1 \cup A_2))=\mu^*(B\cap A_1)+\mu^*(B\cap A_2)$$
Remark: In your attempt, the step

Since $C:=B\cap(A_1\cup A_2)^C=(B\cap A_1^C)\cap A_2^C$, $\mu^*(B\cap(A_1\cup A_2)) + \mu^*(C) = \mu^*(B\cap A_1)+\mu^*(B\cap A_2)+\mu^*(C)$.

is not necessary and it actually leads to an unnecessary more complex path.
