Prove if $A,B\subset\mathbb{R}$ and $|A|<\infty$, then $|B\setminus A|\geq |B|-|A|$ ("Measure, Integration & Real Analysis" by Sheldon Axler)

Question

I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.

2.5 outer measure preserves order on p.16
Suppose $A$ and $B$ are subsets of $\mathbb{R}$ with $A\subset B$. Then $|A|\leq |B|$.

2.8 countable subadditivity of outer measure on p.17
Suppose $A_1,A_2,\dots$ is a sequence of subsets of $\mathbb{R}$. Then $$\left|\bigcup_{k=1}^{\infty}A_k\right|\leq \sum_{k=1}^{\infty} |A_k|.$$

The following exercise is Exercise 3 on p.23 in Exercises 2A in this book:

Exercise 3 on p.23
Prove if $A,B\subset\mathbb{R}$ and $|A|<\infty$, then $|B\setminus A|\geq |B|-|A|$.

My solution is here:

Since $A\cup B=(B\setminus A)\cup A$, $|A\cup B|\leq |B\setminus A|+|A|$ by 2.8 on p.17.
Since $|A|<\infty$, $|A\cup B|-|A|\leq |B\setminus A|$ holds.
Since $B\subset A\cup B$, $|B|\leq |A\cup B|$ by 2.5 on p.16.
Since $|A|<\infty$, $|B|-|A|\leq |A\cup B|-|A|$.
So, $|B|-|A|\leq|A\cup B|- |A|\leq |B\setminus A|$.

Is my solution ok?
Is my solution a standard solution?
I want to know a standard solution for this exercise.

Answer 1

Yes, your solution is correct and it is the standard proof of this inequality. You are basically using 3 facts:

1. Monotonicity of outer measure (2.5 in your textbook): $$B \subset (B \setminus A) \cup A \implies \left|B\right| \leq \left| (B \setminus A) \cup A\right|$$

2. Countable subadditivity of outer measure (2.8): $$ \left| (B \setminus A) \cup A\right| \leq \left| B \setminus A \right| + |A|$$

3. The outer measure of $A$ is finite so that we can subtract it on both sides of the resulting inequality: $$ |B| - |A| \leq |B \setminus A|$$