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.
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.