why Rudin said that let $\Delta$ be closed disc? I have some confusion in RCA book
my confusion is given below and marked in red box

My attempt :  I think Rudin is wrong . Here  it  is given that  $S$ is closed   we know that complement of closed set is open  so $S^c$ is an  open  I think  it should  be let $\Delta$  be open disc because any open ball  is  a countable union of closed ball
My doubt : why Rudin said  that let $\Delta$  be closed  disc ?
 A: It doesn't really matter. First of all, it's true you could use an open disc and you would get $|f-\alpha|<r$, but then you would need to use some integral inequality to complete the proof anyway.
It think, but I'm not entirely sure, that at that point in the book you know that a positive function on a set with positive measure has positive Lebesgue integral. This is: if $g>0$ on a measurable set $A$ such that $\mu(A)>0$, then $\int_A g\,\text{d}\mu>0$. If you know this then you can use $|f-\alpha|<r|\Rightarrow r-|f-\alpha|>0\Rightarrow\int_E (r-|f-\alpha|)\,\text{d}\mu>0\Rightarrow\int_E |f-\alpha|\,\text{d}\mu<\int_E r\,\text{d}\mu$
and get $\left|A_E(f)-\alpha\right|<r$, which would be a contradiction.
Now, on the other hand, I am entirely sure the at that point in the book you know that $g\le h$ on $A$ implies $\int_A g\,\text{d}\mu\le\int_A h\,\text{d}\mu$. In a way, this property is "more fundamental" than the previous one, and that's why I think Rudin wants to use a closed disc, since that gives you $|f-\alpha|\le r$. But, why can he? Because any open set can also be generated as a countable union of closed discs.
Imagine you have an open set $A$ as a countable union of open discs $U_i$, each centred at $\alpha_i$ with radius $r_i$. So $A=\bigcup_i U_i$. Now take one $U_i$ and consider the collection of closed discs $V_{j_i}=\overline B(\alpha_i,r_i-\frac{1}{j})$. Each $V_{j_i}$ is a closed disc, but the countable union $\bigcup\limits_{j=1}^\infty V_{j_i}=\bigcup\limits_{j=1}^\infty \overline B(\alpha_i,r_i-\frac{1}{j})=B(\alpha_i,r_i)=U_i$. So you have each $U_i$ as a countable union of closed discs. Then $A=\bigcup_i U_i=\bigcup_i\big(\bigcup\limits_{j=1}^\infty V_{j_i}\big)$, which is the countable union of countable unions, hence countable. Therefore we have $A$ as a countable union of closed discs.
