# Does the pointwise limit of a sequence {$f_{n}$} of measurable functions imply that f is measurable? [Graduate studies]

I am using the H.L Royden 4th ed and the Folland

Any definition of measurable functions or otherwise is as defined in the text above

1.Let {$$f_{n}$$} be a sequence of measurable functions converging a.e(almost everywhere) to the function $$f$$ on $$E$$. Prove that $$f$$ is measurable.

2.Let ($$\Omega$$, $$\Sigma$$) be a measurable space. If $$f$$ is a pointwise limit of a sequence {$$f_{n}$$} of measurable functions on a common domain $$D$$ $$\in$$ $$\Sigma$$. Then $$f$$ is measurable.

My attempt For 1. I choose a set say $$A_{0}$$ such that this set is a subset of the entire set say $$A$$ . Next i proved that the measure of the first set is 0 and showed that $$f_{n}$$ converges pointwise to the compliment of the first set.

Thereafter we know that f is measurable if and only if the restriction of the set is measurable. I then say that from this I assumed that the convergence to all possible sets A is pointwise.

Then choose an element say b inside the entire real line and used the definition of pointwise convergence so show that f(x) is less than the element say b. And after some working with these sets I showed that the intersection of the entire set belongs to $$M$$. After which the union of my entire set such that b is inside A is equal to my f(x) being less than b which shows that f is measurable since the countable union of measurable sets is again measurable so my LHS is measurable and my RHS is measurable so hence f is measurable.

For 2. I have no idea how to even begin attempting the question. What my thought process?I have never done a question with ($$\Omega$$, $$\Sigma$$). Can anyone point me in the right direction how to begin such a question.

• what is M in the second to last paragraph? – user24142 Oct 26 '18 at 4:57
• If the measure space is not complete (in the sense of measure theory), then the first statement is false, because you can modify $f$ outside the set of convergence such that $f$ is not measurable. (Just take a Vitali-like set!) – p4sch Oct 26 '18 at 6:38