If a f's restriction to D is measurable and E~D is measurable, then f is measurable on E

Question

I am trying to prove the following proposition:

Let f be an extended real-valued function on E. Let D be a measurable subset of E.

1. (i) If f is measurable on E and f = g a.e. on E, then g is measurable on E.

2. (ii) f is measurable on E if and only if it's restriction to D and E~D are measurable.

And I can do it, until I reach 2(ii) going "<-".

How should I go about it? I found this answer: $f$ is measurable iff its restrictions are measurable but the thing is he is only going in one direction and not the other.

I feel like the proof should be trivial because even Royden's skip over it, but I can't figure it out...

Answer 1

It's just that the union of any two measurable sets is always a measurable set. This is a part of the definition of a $\sigma$-algebra!

Given that the restriction of $f$ to $D$ and $E - D$ are measurable, we know, for any $a \in \mathbb R$, that $f^{-1}(a, \infty) \cap D$ and $f^{-1}(a , \infty) \cap (E - D)$ are both measurable sets. But $$f^{-1}(a,\infty) = (f^{-1}(a, \infty) \cap D) \cup (f^{-1}(a , \infty) \cap (E - D)),$$ so $f^{-1}(a,\infty)$ is measurable too. Since $a$ is arbitrary, this proves that $f$ is a measurable function.