Xrfgtjtk

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.