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How to show that $m^*(A \cup B) + m^*(A \cap B) \leq m^*(A)+m^*(B)$ for any $A,B \subseteq \mathbb{R}$.

I first thought that I could easily prove it by using the sub-additivity property of the Lebesgue measure. Soon I realized that it was a bit harder than it seemed because for general sets, the best that we have is $m^*(A)-m^*(B) \leq m^*(A-B) $ provided that $B \subseteq A$ and that led me nowhere. So, I have decided to prove it using the definition because the property given in the problem is obviously true for intervals, but I’m still having trouble writing it down because if I want to do it with double-sums I get nonsensical equations:

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$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}l(A_i \cup B_j) = \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}l(A_i)+ \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}l(B_j) – \sum_{i=1}^{\infty}\sum_{j=1}^{\infty} (A_i \cap B_j)$$

which is absurd because the first two sums in the R.H.S are infinite while the L.H.S can be finite.

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I’m going to post an answer to provide a proof of the above-mentioned problem because it may be handy in the future for other people:

Since $m^*(A)=\inf\{m(O): A \subseteq O\}$, for any $\epsilon>0$ we find $A \subseteq O_1$ and $B\subseteq O_2$ such that:

$$ m(O_1) < m^*(A)+ \epsilon/2 $$

$$ m(O_2) < m^*(B)+ \epsilon/2 $$

$$ m(O_1) + m(O_2) < m^*(A) + m^*(B)+ \epsilon $$

Open sets are measurable and for measurable sets we have the equality: $$m(O_1) + m(O_2) = m(O_1 \cap O_2) + m(O_1 \cup O_2)$$

Therefore, we get:

$$ m(O_1 \cap O_2) + m(O_1 \cup O_2) < m^*(A) + m^*(B)+ \epsilon $$

But by definition of the Lebesgue measure, noting that $A\cap B \subseteq O_1 \cap O_2$ and $A\cup B \subseteq O_1 \cup O_2$, we get:

$$m^*(A \cap B) + m^*(A \cup B) \leq m(O_1 \cap O_2) + m(O_1 \cup O_2)$$

Which gives:

$$m^*(A \cap B) + m^*(A \cup B) < m^*(A) + m^*(B)+ \epsilon$$

Since $\epsilon$ was an arbitrary positive number, we finally obtain:

$$m^*(A \cap B) + m^*(A \cup B) \leq m^*(A) + m^*(B)$$

(Thanks to Daniel Fischer for his key idea)

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