# Property of Outer Measure on $\mathbb{R}$

Question From - Axler Measure Theory - Problem 3 - Section 2A

Throughout: For $$A \subset \mathbb{R},$$ $$|A|$$ denotes the outer measure of $$A$$ and is defined

$$|A|=inf\\{\sum_{k=1}^{\infty}\ell(I_k): I_1, I_2, \cdots \text{open intervals}, A \subset \cup_{k=1}^{\infty} I_k\\}$$

Let $$A,B \subset \mathbb{R}$$ with $$|A|< \infty.$$ Show that $$|B|-|A| \leq |B \setminus A|$$

I was able to show the result for when $$A \subset B$$ because in this case we have

$$|B|=|(B \setminus A) \cup A| \leq |B \setminus A|+|A|$$

However, the general case I am unable to show. I really think it is just a matter of the right set identity and properties of outer measures, but I cannot find the right one. I am aware that $$B \setminus A = B \cap A^c.$$ Thank you for the help.

• $B = (B \cap A) \cup B \setminus A \subset A \cup B \setminus A$. Commented Mar 13 at 17:32

This is a good start. You've shown that you have the desired inequality when $$A \subseteq B$$, so next you should ask yourself "is the case $$A \nsubseteq B$$ any harder?"
Imagine we start with $$A \subseteq B$$, and then enlarge $$A$$ by adding elements from $$B^c$$ to get a set $$A'$$ with $$A \subseteq A' \nsubseteq B$$. Because we only added elements which were not in $$B$$, we have $$B \setminus A' = B \setminus A$$, so $$\lvert B \setminus A' \rvert = \lvert B \setminus A \rvert$$. On the other hand, $$\lvert A' \rvert \geq \lvert A \rvert$$, so we get $$\lvert B \rvert - \lvert A' \rvert \leq \lvert B \rvert - \lvert A \rvert$$. Overall, we get $$\lvert B \rvert - \lvert A' \rvert \leq \lvert B \rvert - \lvert A \rvert \leq \lvert B \setminus A \rvert = \lvert B \setminus A' \rvert$$.
In other words, adding elements from $$B^c$$ to $$A$$ makes the inequality easier to show! The other good news is that every subset of $$\mathbb{R}$$ can be produced by starting with a subset of $$B$$ and then adding in elements of $$B^c$$. So, you don't need to produce any new clever argument. You just need to use this observation to extend your result from the case $$A \subseteq B$$ to the general case.