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

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    $\begingroup$ Use the monotonicity of outer-measure. $\endgroup$
    – FreeMind
    Commented Feb 14, 2021 at 12:17
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    $\begingroup$ I have posted a detailed answer to your question. Please, let me know if you have any question regarding my answer. If my answer provides relevant / helpful information regarding your question, please, upvote it. If my answer actualy answers your question, accept it too, please. To upvote, click the triangle pointing upward above the number (of votes) in front of the question. To accept the answer, click on the check mark beside the answer to toggle it from greyed out to filled in. $\endgroup$
    – Ramiro
    Commented Feb 14, 2021 at 13:33
  • $\begingroup$ @FreeMind Thank you very much for your comment. $\endgroup$
    – tchappy ha
    Commented Feb 15, 2021 at 0:54
  • $\begingroup$ @Ramiro Thank you very much for your answer. $\endgroup$
    – tchappy ha
    Commented Feb 15, 2021 at 0:54

2 Answers 2

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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.

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  • $\begingroup$ Ramiro, Thank you very much for your answer. $\endgroup$
    – tchappy ha
    Commented Feb 15, 2021 at 0:55
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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 $$C=(B\cap(A_1\cup A_2))\cap(A_1\cup A_2)^C=\emptyset.$$

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  • $\begingroup$ Liufeng Yang, Thank you very much for your answer. $\endgroup$
    – tchappy ha
    Commented Feb 15, 2021 at 0:54

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