# Additivity of Outer Measure for Separated Bounded Sets (using definition of an outer measure only)

Suppose $A$ and $B$ are separated, bounded sets; i.e., there is an $\alpha \in \mathbb{R}$, $\alpha > 0$ such that $|a - b|\geq \alpha$ for all $a \in A$, $b \in B$. Prove $\underline{\mathbf{\text{directly from the definition}}}$ of $\mathbf{m^{*}}$ that $\mathbf{m^{*}(A \cup B) = m^{*}(A)+m^{*}(B)}$.

This question has been asked before on here, although not exactly in this way. Further, none of the answers have been sufficient - just hints that don't do enough at 1) including intermediate steps and 2) explaining the why behind those intermediate steps.

I would like a full solution (with proper subscripting - I'm having some issues with that in my own solution) using the same method (showing directly by the definition of an outer measure $m^{*}$).

I have made several attempts at this problem. This is the gist of them:

1. $\underline{\text{Showing that}\, \mathbf{m^{*}(A \cup B) \leq m^{*}(A)+m^{*}(B)}}$: Consider $I = \left( a-\frac{\alpha}{4}, a + \frac{\alpha}{4} \right)$ and $J=\left(b-\frac{\alpha}{4},b+\frac{\alpha}{4} \right)$, where $a \in A$, $b \in B$. Then, $A \subseteq \cup_{a \in A}I$, but $\left(\cup_{a \in A}I \right)\cap B = \emptyset$ since each $l(I) = \frac{\alpha}{4} < \alpha \leq |a-b|$ (where $l$ is the length of the interval).

Similarly, we have $B \subseteq \cup_{b \in B}J$, where $\left(\cup_{a \in A}J \right)\cap A = \emptyset$ since each $l(J) = \frac{\alpha}{4} < \alpha \leq |a-b|$.

So, $A \cup B \subseteq \left[\cup_{a \in A}I \right] \cup \left[\cup_{b \in B}J \right] = \cup_{c \in A\cup B}K$.

Finally, since $m^{*}$ is a measure, $\forall \epsilon > 0$, we have $m^{*}(A \cup B) = \inf\left( \sum_{c \in A \cup B} l(K_{c})\right)= \inf \left( \sum_{a \in A}l(I_{a}) + \sum_{b \in B}l(J_{b})\right)$ (since $A$ and $B$ are disjoint, we can consider separately those intervals covering $A$ from those covering $B$) $\leq \sum_{a \in A}l(I_{a})+\sum_{b \in B}l(J_{b})\leq m^{*}(A)+\frac{\epsilon}{2} + m^{*}(B) + \frac{\epsilon}{2} = m^{*}(A) + m^{*}(B)+\epsilon$.

Since this holds for all $\epsilon > 0$, it holds for $\epsilon = 0$

2. $\underline{\text{Showing that}\, \mathbf{m^{*}(A)+m^{*}(B) \leq m^{*}(A \cup B)}}$: Although I know there are problems with the other direction, I feel completely lost in this direction.

Using the same covers for $A$ and $B$ from the first direction, by definition of the outer measure, we have $m^{*}(A) + m^{*}(B) = \inf \left\{\sum_{n=1}^{\infty}l(I_{n}) \right\}+\inf \left\{\sum_{n=1}^{\infty}l(J_{n}) \right\}$

Now, from here I am not sure how to proceed. I suspect that since as we add in more and more terms, the sums get bigger, that the two greatest lower bounds ($\inf$s) each equal $\frac{\alpha}{4}$. Then the above inequality would in turn be $=\frac{\alpha}{2} < \alpha \leq |a-b|$, but I'm not entirely sure how that would help me.

Essentially, what I would like to see someone post in a solution is a full solution using the definition of the outer measure $m^{*}$. What would be ideal is a solution based on what I have done here with all the bad bits fixed.

## 1 Answer

First, if your definition of $m^*$ is the axiomatic one (see here), then part 1. of your proof is unnecessary because outer measures are countably subadditive by definition. But since you are working with Lebesgue measure on $\mathbb{R}$ I will assume your definition is $$m^*(A) = \inf\{|U|: U\supset A, ~U~\text{open}\}$$ where $|U|$ denotes Lebesgue measure or premeasure, which should be well-defined at this point for open sets.

Then for part 2. what we want to show is that $m^*(A)+m^*(B)$ is a lower bound for the set $$\{|U|: U\supset A\cup B,~U~\text{open}\}.$$ Ok, so let's take an open set $U$ containing $A\cup B$. Also, let's cover $A$ with balls of radius $a/4$, and call the union of these balls $U_A$. Similarly, cover $B$ with balls of radius $a/4$ and call the union $U_B$. Then $U_A$ and $U_B$ are both open and disjoint by the separation property. Then $V_A = U_A\cap U$ and $V_B = U_B\cap U$ are also open, still disjoint, and $V_A\supset A$, $V_B\supset B$. By definition of outer measure, $$m^*(A)\leq |V_A|<\infty,~m^*(B)\leq |V_B|<\infty.$$ The finiteness of these quantities is due to the assumption that $A$ and $B$ are bounded. We then observe that since $V_A$ and $V_B$ are pairwise disjoint and open, $$|V_A| + |V_B| = |V_A\cup V_B|.$$ Since $V_A\cup V_B\subset U$, it follows now that $$m^*(A)+m^*(B) \leq |V_A| + |V_B| = |V_A\cup V_B| \leq |U|.$$ (Notice what we have in the middle - we've moved ourselves from the realm of arbitrary sets to the realm of open sets, where we know how Lebesgue premeasure behaves. This is the key idea.) Since $U$ was an arbitrary open set containing $A\cup B$, it then follows by definition of infimum that $$m^*(A)+m^*(B) \leq m^*(A\cup B).$$

• is my part 1 correct, though? You didn't say anything about it. – ALannister Oct 12 '15 at 5:10
• While I suggest you write it up more carefully than as it is right now, the ideas for your part 1 are generally sound. – Gyu Eun Lee Oct 13 '15 at 2:52
• the definition of outer measure I'm using is $m^{*}(A)=\sum_{k=1}^{\infty}l(I_{k})$, where $A\subseteq\cup_{k=1}^{\infty}I_{k}$. How do I express $V_{A}$ and $V_{B}$ this way so I can use definition of outer measure to say $m^{*}(A) \leq \sum_{k=1}^{\infty}(\text{intervals in}\,V_{A})$, $m^{*}(B) \leq \sum_{k=1}^{\infty}(\text{intervals in}\,V_{B})$, and also have $\sum_{k=1}^{\infty}(\text{intervals in}\,V_{A}) + \sum_{k=1}^{\infty}(\text{intervals in}\,V_{B}) = \sum_{k=1}^{\infty}(\text{intervals in}\,V_{A}\cup V_{B})$ from the fact that $V_{A}$ and $V_{B}$ are pairwise disjoint? – ALannister Oct 13 '15 at 8:27
• In one dimension, open balls are disjoint unions of at most countably many intervals. – Gyu Eun Lee Oct 13 '15 at 20:34