Q: why $E-F=W-E^c?$ From Rudin book Real and complex analysis ,page no : $48$

$\bf 2.17\ $ Theorem $\ $ Suppose $X$ is a locally compact, $\sigma$-compact Hausdorff space. If $\frak M$ and $\mu$ are as described in the statement of Theorem $\it 2.14$, then $\frak M$ and $\mu$ have the following properties:
$(a)\ \ $ If $E\in\frak M$ and $\epsilon>0$, there is a closed set $F$ and an open set $V$ such that $F\subset E\subset V$ and $\mu(V-F)<\epsilon$.
$(b)\ \ $ $\mu$ is a regular Borel measure on $X$.
$(c)\ \ $ If $E\in\frak M$, there are sets $A$ and $B$ such that $A$ is an $F_\sigma$, $B$ is a $G_\delta$, $A\subset E\subset B$, and $\mu(B-A)=0.$

In the third paragraph proof  Rudin say that apply this to $E^c$ in place of $E$ :There is an open set  $W \supset E^c $ such that $\mu(W-E^c)  < \epsilon /2.$If $F= W^c $, then $F \subset E$ and $E-F=W-E^c$. Now $(a)$ follow
Im confused that  why $E-F=W-E^c?$
Here $F=W^c$ so  it should be  $E-F=E-W^c $

A: $p \in E - W^c$ if and only if it is in $E$ and it is not in $W^c$. Not being in $W^c$ is the same as being $W$. I.e., $$E - W^c = E \cap W$$
I'll assume I don't need to go further.
