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Denote the distance between two sets $A,B \in \Bbb R^n$ as $d(A,B).$ If $d(A,B) > 0$ show that $m^*(A \cup B) = m^*(A) + m^*(B)$.

The part $\le$ seems to come from subaddtivity of outer measure. That is $$m^*(A \cup B) \le m^*(A) + m^*(B).$$

Now if $\{I_k \}_k$ is a cover for $A$, then $m^*(A) \le \sum_{k=1}^\infty \ell(I_k)$ similarly if $\{J_k\}_k$ is a cover for $B$, then $m^*(B) \le \sum_{k=1}^\infty \ell(J_k)$.

Now if $\{S_k\}_k$ is a cover for $A\cup B$ then from the definition of infimum I have that $m^*(A \cup B) + \varepsilon\ge \sum_{k=1}^\infty \ell(S_k)$.

So what I have is that $$m^*(A) + m^*(B) \le \sum_{k=1}^\infty (\ell(I_k)+\ell(J_k))$$

and it seems that I would somehow need to connect this with $m^*(A \cup B) + \varepsilon\ge \sum_{k=1}^\infty \ell(S_k) $. Any hints on what should I do here?

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3 Answers 3

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An alternative approach: First show that if $O$ is open, then $O = \bigcup_{i = 1}^{\infty}Q_i$, where the $Q_i$ are dyadic cubes disjoint except for possibly their boundaries. Then show $m^*(O) = \sum_{i = 1}^{\infty}\ell(Q_i)$. Deduce that for any $S \subset \mathbb{R}^n$, $m^*(S) = \inf\{m^*(O) : O \text{ open }, O \supset S\}$. Also we easily see via the dyadic cube decomposition that if $O_1$ and $O_2$ are disjoint, then $m^*(O_1 \cup O_2) = m^*(O_1) + m^*(O_2)$.

Using the above, the proof that $m^*(A) + m^*(B) \leq m^*(A \cup B)$ is not difficult.

Let $O$ be an arbitrary open set containing $A \cup B$. Let $r = d(A, B) > 0$. Let $O_1 = O \cap \{x \in \mathbb{R}^n : d(x, A) < r/2\}$, $O_2 = O \cap \{x \in \mathbb{R}^n : d(x, B) < r/2\}$. Then \begin{align} m^*(A) + m^*(B) &\leq m^*(O_1) + m^*(O_2) \\ &= m^*(O_1 \cup O_2) \\ &\leq m^*(O). \end{align} Since $O$ was arbitrary, $m^*(A) + m^*(B) \leq m^*(A \cup B)$.

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Hint: fix $\varepsilon > 0$ and pick a cover $(S_k)$ of $A \cup B$ so that $$\sum_{k=1}^{\infty} \ell(S_k) < m^*(A \cup B) + \varepsilon.$$

Now prove that without loss of generality we can assume that $(S_k)$ consists of rectangles of diameter smaller than $d(A, B)$. Then partition the cover into two parts so that the first one will cover $A$ and the second will cover $B$.

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  • $\begingroup$ So if $S_k = \bigcup_{j} I_j$, then for each $I_j$, we have that $\ell(I_j) \le d(A,B)?$ $\endgroup$
    – Georgi
    Commented Sep 19, 2021 at 19:14
  • $\begingroup$ If you're asking whether the implication you wrote is always true, then of course it is not always true. Otherwise I don't know what you're asking. $\endgroup$
    – Adayah
    Commented Sep 20, 2021 at 10:29
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Call a cover $\{I_k\}$ of $A$ an $\epsilon$-box cover if all boxes have all dimensions $< \epsilon$, and where all boxes have a nonempty intersection with $A$.

Let's first acknowledge that for any $\epsilon > 0$, $m^*(A)$ can be expressed as $\inf \sum\limits_{k = 1}^\infty \ell(I_k)$, where the infimum is taken specifically over $\epsilon$-box covers $\{I_k\}$.

We can do this because given a box $I$ with the largest dimension being $J \geq \epsilon$, we can pick $n > \frac{J}{\epsilon}$ and subdivide the box into $n$ boxes with that dimension being $\frac{J}{n} < \epsilon$. We can then throw away any boxes that do not intersect $A$ from the cover to get a possibly smaller cover.

Now let $\epsilon = \frac{d(A, B)}{2n}$. Consider some $\epsilon$ box cover $\{I_k\}$ of $A \cup B$. Then note that for no box $I_j$ does $I_j$ intersect both $A$ and $B$, since for any two points $a, b \in I_j$, $d(a, b)^2 \leq n (\frac{d(A, B)}{2n})^2 = \frac{d(A, B)}{4n} < d(A, B)$. So we can divide the $\epsilon$ box cover $\{I_k\}$ of $A \cup B$ into an $\epsilon$-box cover of $A$, plus an $\epsilon$-box cover for $B$, since for all $j$, exactly one of $I_j \cap A \neq \emptyset$ and $I_j \cap B \neq \emptyset$ is true.

Conversely, if we have an $\epsilon$-box cover of $A$ and an $\epsilon$-box cover of $B$, we can put them together to get an $\epsilon$-box cover of $A \cup B$.

Therefore, $m^*(A \cup B) = m^*(A) + m^*(B)$.

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