Show that the family $A^*$ of $\mu^*$-measurable sets is $\sigma$-algebra (proof verification)

Question

This is from Bartle text book, and this page proves that the family $A^*$ of $\mu^*$-measurable sets is $\sigma$-algebra. I don't understand the red line. Could you elaborate on this?

Answer 1

The simplest is just to draw a Venn diagram of $E,A,F$ and shade the areas involved.

$B$ is defined as $A \setminus (E \cap F)$. Show that $B \cap F = (A \cap F) \setminus E$ by two inclusions:

Suppose $x \in B \cap F$, then $x \in F$ and $x \in B$, which means $x \in A$ and $x \notin E \cap F$. We already know $x \in F$ so $x \notin E \cap F$ implies that $x \notin E$. So $x \in A \cap F$ and $x \notin E$ so $x \in (A \cap F)\setminus E$.

Suppose that $x \in (A \cap F) \setminus F$ then $x \in A$, $x \in F$ and $x \notin E$. So $x \notin E \cap F$ and as we also have $x \in A$ we know that $x \in B$ and we already knew $x \in F$ so that $x \in B \cap F$.

Now $B \setminus F = A \setminus F$ can be seen in a similar way:

$x \in B \setminus F$, then $x \in B$ and $x \notin F$. As $B \subseteq A$ we know that $x \in A \setminus F$.

If $x \in A \setminus F$, then $x \notin E \cap F$ so $x \in A \setminus (E \cap F) = B$. Hence $x \in B \setminus F$.