I have a question about this proof of a theorem from Yeh’s Real Analysis (3rd edition).
The theorem and proof are in the attached images:
My question pertains to the part that says:
$\{D : g < r_n\} = \{D \setminus N : g < r_n\} \cup \{N : g < r_n\}$
How do we get that $D \setminus N \cup N = D$? This is only true if $N \subseteq D$, but I’m not sure exactly how this came to be. I’m guessing it’s from the fact that D is a complete measure space? But it’s a bit vague to me.
Thanks!
Although your question pertains to equation $(1)$, observe that the same assumption, namely that $D = (D \setminus N) \cup N$, is made earlier when defining $g$ on $D$ by defining it separately on $D \setminus N$ and $N$.
You are right that the author is assuming that $N \subset D$, and that this is not true from the way $N$ is defined. However, if we let $N' := N \cap D$, then the same proof goes through with $N'$ in place of $N$. For instance, it is clear that $N' \in \mathfrak{A}$, $N' \subset D$, $N'$ is a null set and $C_n \subset B_n \subset N'$. Note that you are not using the fact that $D$ is a complete measure space, you only need that $D, N \in \mathfrak{A}$ to conclude that $D \cap N \in \mathfrak{A}$, etc.
So, we can take without loss of generality that $N$ is a subset of $D$. But the author should have mentioned this for the sake of clarity, in my opinion. Good catch.