# Lebesgue outer measure of disjoint sets in $\mathbb{R}^n$

If $d(A, B) > 0$, then it's true that $m^*(A\cup B) = m^*(A) + m^*(B)$. If there are disjoint open sets $U, V$ such that $A \subset U$ and $B \subset V$, doesn't it still hold that $m^*(A\cup B) = m^*(A) + m^*(B)$, even in the case $d(A, B) = 0$?

My thinking:

Since subadditivity gives the other direction, we just need to show $m^*(A\cup B) \geq m^*(A) + m^*(B)$. For some $\epsilon>0$, we have a covering of $A\cup B$ by disjoint closed cubes $\bigcup_{j=1}^{\infty} Q_j$, so that $\sum_{i=1}^{\infty} Q_j < m^*(A\cup B) + \epsilon$. If the cubes can be separated into those lying inside $U$ or $V$, we're done. In the case where some cubes intersect both $U$ and $V$, then for such $Q_j$, we can split the interior of $Q_j \cap U$ and $Q_j \cap V$ into coutably many disjoint open cubes. But then we'd have to worry about the boundaries of these sets, which I $think$ we can cover with cubes whose volumes totals no more than $\frac{\epsilon}{4^j}$. Now we are left with disjoint covers of $A$ and $B$ that are very close in volume to $\bigcup_{j=1}^{\infty} Q_j$.

Is this right? If so, is there a slicker proof?

HINT: For every $A$ subset of $\mathbb{R}^n$ we have $$m^{\star}(A)= \inf_{U \textrm{ open } \supset A} m(U)$$

$\bf{Added:}$ In fact, the above statement implies that for every $A$ subset of $\mathbb{R}^n$ we have $$m^{\star}(A)= \inf_{U \textrm{ measurable } \supset A} m(U)$$ Moreover, the equality is achieved for some $U \supset A$, $U$ countable intersection of open sets containing $A$. Therefore, if $A \subset U$, $B\subset V$, and $U$, $V$ are disjoint and ${ \it measurable}$ subsets then $m^*(A\cup B) = m^*(A) + m^*(B)$. Indeed: take $W\supset A \cup B$ measurable so that $m^*(A\cup B) = \mu(W)$. However, $\mu(W) \ge \mu( W \cap (U\cup V) ) = \mu(W \cap U) + \mu ( W \cap V) \ge m^*(A) + m^*(B)$.

WLOG $m^*(A), m^*(B) < \infty$, otherwise both sides are infinite.

Choose open $W_n\supseteq A \cup B$ decreasing with $m(W_n) \searrow m^*(A \cup B)$. Then $W_n \cap U \supseteq A$ and $W_n \cap V \supseteq B$. Thus $$m^*(A \cup B) = \lim_n m(W_n) \geq \limsup_n [m(W_n \cap U) + m(W_n \cap V)]\geq m^*(A) + m^*(B).$$ Of course the other inequality is given immediately by subadditivity.

Your thinking is OK and it can be developed into a complete proof.

Suppose there are disjoint open sets $U, V$ such that $A \subset U$ and $B \subset V$. For each $x\in A$, let $C_x$ be an open cube such that $x\in C_x$ and $\overline{C_x}\subseteq U$. We have that $\{C_x \,|\, x\in A\}$ is open cover of $A$. Since $\mathbb{R}^n$ is separable, there is a countable sub-cover $\{C_{x_n} \,|\, x_n\in A, n\in \mathbb{N} \}$. Note that $$A\subseteq \bigcup_{n\in \mathbb{N}} C_{x_n} \subseteq \bigcup_{n\in \mathbb{N}}\overline{ C_{x_n}} \subseteq U$$

In a similar way, we can find a countable family $\{D_{y_n} \,|\, y_n\in B, n\in \mathbb{N} \}$ of open cubes such that $$B\subseteq \bigcup_{n\in \mathbb{N}} D_{y_n} \subseteq \bigcup_{n\in \mathbb{N}}\overline{ D_{y_n}} \subseteq V$$

Since $U$ e $V$ are disjoint, we can easily use $\{C_{x_n} \,|\, x_n\in A, n\in \mathbb{N} \}$ and $\{D_{y_n} \,|\, y_n\in B, n\in \mathbb{N} \}$ (or, if you prefer, $\{\overline {C_{x_n}} \,|\, x_n\in A, n\in \mathbb{N} \}$ and $\{\overline{D_{y_n}} \,|\, y_n\in B, n\in \mathbb{N} \}$) to prove that $$m^*(A\cup B) = m^*(A) + m^*(B)$$

In fact, let us show the details.

From subadditivity, we know that $m^*(A\cup B) \leq m^*(A) + m^*(B)$. we just need to show $m^*(A\cup B) \geq m^*(A) + m^*(B)$.

If $m^*(A\cup B) =+\infty$ then it is clear that $m^*(A\cup B) \geq m^*(A) + m^*(B)$.

Suppose $m^*(A\cup B) <+\infty$. For any $\epsilon>0$, there is a countable collection of cubes $\{Q_j\}_{j\in\mathbb{N}}$, such that $A\cup B\subseteq \bigcup_{j=0}^{\infty} Q_j$ and $$\sum_{i=0}^{\infty} m(Q_j) < m^*(A\cup B) + \epsilon \tag{1}$$

Since the family of cubes in $\mathbb{R}^n$ is a semi-ring, then from $\{C_{x_n} \,|\, x_n\in A, n\in \mathbb{N} \}$, we can have $\{E_k \,|\, k\in \mathbb{N} \}$ a collection of disjoint cubes, such that $A\subseteq \bigcup_{n\in \mathbb{N}} C_{x_n}=\sum_{k\in \mathbb{N}} E_k$. And from $\{D_{y_n} \,|\, y_n\in B, n\in \mathbb{N} \}$, we can have $\{F_l \,|\, l\in \mathbb{N} \}$ a collection of disjoint cubes, such that $B\subseteq \bigcup_{n\in \mathbb{N}} D_{y_n}=\sum_{l\in \mathbb{N}} F_l$. Note that, for all $k,l \in \mathbb{N}$, $E_k\subseteq U$ e $F_l\subseteq V$, so $E_k \cap F_l=\emptyset$.

Now, we have $A\subseteq \sum_{j,k=0}^\infty(Q_j\cap E_k)$ and $$m^*(A)\leq \sum_{j,k=0}^\infty m(Q_j\cap E_k) \tag{2}$$ (remember that the inersection of two cubes is a cube).

And we also have $B\subseteq \sum_{j,l=0}^\infty(Q_j\cap F_l)$ and $$m^*(B)\leq \sum_{j,l=0}^\infty m(Q_j\cap F_l) \tag{3}$$

Combining $(1)$, $(2)$ and $(3)$, we have $$m^*(A)+m^*(B)\leq \sum_{j,k=0}^\infty m(Q_j\cap E_k)+ \sum_{j,l=0}^\infty m(Q_j\cap F_l)\leq\sum_{i=0}^{\infty} m(Q_j) < m^*(A\cup B) + \epsilon$$ Since this is valid for an arbitrary $\epsilon>0$, we have $$m^*(A)+m^*(B)\leq m^*(A\cup B)$$

Remark: IF you already know that, for all $A$ subset of $\mathbb{R}^n$, we have $$m^{\star}(A)= \inf_{U \textrm{ open } \supset A} m(U)$$ THEN there are much shorter and elegant proofs (see, for instance, orangeskid's answer).

• so is the idea that if we have a really tight covering of $A \cup B$ by closed cubes, we can take the intersection of that covering with the $C_{x_n}$'s or $D_{x_n}$'s to get a covering of A and B as well? Commented Aug 14, 2015 at 19:06
• Sorry, I guess the easy step that you mentioned is not entirely clear to me Commented Aug 14, 2015 at 19:20
• @dddmsj Yes, if you have a covering of $A\cup B$ by closed cubes, you take the intersection of that covering with the $C_{x_n}$'s to get a covering of $A$ and and the intersection of that covering with $D_{y_n}$ 's to get a covering of $B$. And the union of those two coverings will a covering "finer" than the original covering of $A\cup B$. And so we can prove that $m^*(A\cup B)\geq m^*(A)+m^*(B)$. Commented Aug 14, 2015 at 21:13
• @dddmsj I have edited my answer and included the details. Commented Aug 14, 2015 at 23:22
• thank you for elaborating. I think the part that is hazy to me is why the intersection of two countable union of cubes can also be expressed as a countable union of cubes. However, I think my whole line of tinkering can be avoided if I just thought of open sets instead. Commented Aug 15, 2015 at 6:04