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Lemma 7.5.3: Let $\Omega$ be a measurable subset of $\mathbb{R^n}$ , and let $f: \Omega \to \mathbb{R}^m$ be a function. Then $f$ is measurable iff $f^{-1}(B)$ is measurable for every open box $B$.

This definition makes me feel that measurability is really similar to continuity in the sense that both involve inverse of some subset with some property be subset of similar property (be it open or measurable). This makes me feel that that measurability and continuity are somehow similar but somehow different at the same time.

Could it be explained why these conceptually different properties look so similar?

tryst with freedom
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    Continuity implies (Borel, hence Lebesgue) measurability, almost by definition. These notions are similar in the way the definitions are phrased, but different in the class of sets involved. The (Borel) sigma-algebras are much larger than the topologies hence measurability is a weaker notion of regulaity. And generally many properties are phrased in terms of preimages because preimages behave well. I'm sure a category-theoretic explanation can be given here but that's out of my expertise. – peek-a-boo Aug 01 '22 at 12:03
  • ". The (Borel) sigma-algebras are much larger than the topologies hence measurability is a weaker notion of regulaity." @peek-a-boo could you elaborate this one – tryst with freedom Aug 01 '22 at 20:43
  • which part do you need elaboration on? – peek-a-boo Aug 01 '22 at 20:58
  • I quoted it,.. @peek-a-boo – tryst with freedom Aug 01 '22 at 20:59
  • There's 3 things in that sentence: 1) Borel sigma-algebra is larger. 2) measurability is a weaker notion than continuity (follows from (1)). 3) the notion of regularity... which is very much related to point (2). So, which part do you need elaboration on? – peek-a-boo Aug 01 '22 at 21:00
  • Related: https://math.stackexchange.com/q/1506719/169085 – Alp Uzman Aug 01 '22 at 21:00
  • @peek-a-boo Not OP, but it's not obvious to me that a larger set of sets leads to weaker regularity. Sure there are more places the preimages can land, but there are also more things that have to land in the right spot. – Charles Hudgins Aug 02 '22 at 01:52
  • @CharlesHudgins sure, but when we consider topological-space valued functions (e.g $\Bbb{R}$ or $\Bbb{R}^n$ or $\Bbb{C}$) the $\sigma$-algebra on the target is almost by default the Borel $\sigma$-algebra, for which it is equivalent to only check that preimages of open sets land in the right spot. OP essentially quoted this result which is why I made the comment about regularity. – peek-a-boo Aug 02 '22 at 02:09
  • @peek-a-boo yeah that makes sense. Essentially relying on how well-behaved pre-images are with respect to set operations and that open sets are the building blocks of measurable sets. – Charles Hudgins Aug 02 '22 at 02:11
  • Why are borel algebra bigger than topologies generally ? @peek-a-boo – tryst with freedom Aug 02 '22 at 12:46
  • @Beautifullyirrational google the definition of Borel $\sigma$-algebra and tell me. – peek-a-boo Aug 02 '22 at 13:34

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