This is about an exercise in [Bass]:
Exercise 19.5. Prove that if is infinte-dimensional, that is, it has no finite basis, then the closed unit ball in is not compact.
Proof. Choose an orthonormal basis , then . This means the sequence is not Cauchy hence has no convergent subsequence.
For a Banach space, by Riesz’s lemma to find a non-Cauchy sequence.
[Bass] Bass, R. F. (2013). Real analysis for graduate students. Createspace Ind Pub.
Here we give proofs for two versions of Lusin’s Theorem, one from Exercise 44, Ch2 in Folland’s Real Analysis and the other from the textbook used for my first year undergraduate mathematical analysis course in Beijing. The latter version is a stronger result which in addition discusses the condition for a real-valued function defined on a subset of to be extended to the whole of . A more general result in topology is the Tietze Extension Theorem.
See the full post here: Lusin’s Theorem and Continuous Extension
Here we let denote the Lebesgue measure on .
Lusin’s Theorem (Version 1)[Exercise 2.44, Folland]. Suppose is Lebesgue measurable, is Lebesgue measurable and , there is a compact set such that and is continuous.
Lusin’s Theorem(Version 2)[Huan]. Suppose is Lebesgue measurable and is a Lebesgue measurable extended real valued function with , then , such that , where denotes the space of continuous function on .
Continuous Extension Theorem[Huan]. Suppose , then can be extended to a continuous function on if and only if can be extended to a continuous function on the closure of .
Tietze Extension Theorem. Let be normal and be closed and let be continuous. Then there is a map such that
for all . (Note that in topology, by a map we mean a continuous function. )