I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
Let $A$ be a Borel subset of $\mathbb{R}.$
Let $B$ be a Borel subset of $\mathbb{R}.$
Suppose $A\cap B=\emptyset.$
Then, $|A\cup B|=|A|+|B|$ holds.
Let $A'$ be a subset of $\mathbb{R}.$
Let $B'$ be a subset of $\mathbb{R}.$
Then, $|A'\cup B'|\leq |A'|+|B'|$ holds.
$C$ is a closed set such that $C\subset\mathbb{R}.$
$E$ is a Borel subset of $\mathbb{R}.$
$F$ is a closed set such that $F\subset C\cap E.$
The author used the following inequality when the author proved Luzin's Theorem, second version.
$|((C\cap E)\setminus F)\cup (\mathbb{R}\setminus C)|\leq |(C\cap E)\setminus F|+|\mathbb{R}\setminus C|.$
On the other hand,
$(C\cap E)\setminus F$ is a Borel subset of $\mathbb{R}.$
$\mathbb{R}\setminus C$ is a Borel subset of $\mathbb{R}.$
And $((C\cap E)\setminus F)\cap (\mathbb{R}\setminus C)=\emptyset.$
So, $|((C\cap E)\setminus F)\cup (\mathbb{R}\setminus C)|= |(C\cap E)\setminus F|+|\mathbb{R}\setminus C|$ holds.
I think the author used $$|((C\cap E)\setminus F)\cup (\mathbb{R}\setminus C)|\leq |(C\cap E)\setminus F|+|\mathbb{R}\setminus C|$$ because this inequality is sufficient for his proof.
But $|((C\cap E)\setminus F)\cup (\mathbb{R}\setminus C)|= |(C\cap E)\setminus F|+|\mathbb{R}\setminus C|$ contains more information than $|((C\cap E)\setminus F)\cup (\mathbb{R}\setminus C)|\leq |(C\cap E)\setminus F|+|\mathbb{R}\setminus C|.$
In this situation, which is better to use $$|((C\cap E)\setminus F)\cup (\mathbb{R}\setminus C)|\leq |(C\cap E)\setminus F|+|\mathbb{R}\setminus C|$$ or $$|((C\cap E)\setminus F)\cup (\mathbb{R}\setminus C)|= |(C\cap E)\setminus F|+|\mathbb{R}\setminus C|$$?