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

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?

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

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 $$B$$.

• So if $S_k = \bigcup_{j} I_j$, then for each $I_j$, we have that $\ell(I_j) \le d(A,B)?$ Commented Sep 19, 2021 at 19:14
• 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. Commented Sep 20, 2021 at 10:29

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