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 $$\mu^*(C)=\infty$$.

• Use the monotonicity of outer-measure. Commented Feb 14, 2021 at 12:17
• @FreeMind Thank you very much for your comment. Commented Feb 15, 2021 at 0:54
• @Ramiro Thank you very much for your answer. Commented Feb 15, 2021 at 0:54

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.

• Ramiro, Thank you very much for your answer. Commented Feb 15, 2021 at 0:55

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:

1. 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;

2. 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 $$C=(B\cap(A_1\cup A_2))\cap(A_1\cup A_2)^C=\emptyset.$$

• Liufeng Yang, Thank you very much for your answer. Commented Feb 15, 2021 at 0:54