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I am interested in the way the different generalized definitions of measurable functions relate to each other. Here are a few I am considering:

From Folland

If $(X, \mathcal{M})$ and $(Y, \mathcal{N})$ are measurable spaces, a mapping $f: X \to Y$ is called $(\mathcal{M}, \mathcal{N})$-measurable, or just measurable when $\mathcal{M}$ and $\mathcal{N}$ are understood, if $f^{-1}(E) \in \mathcal{M}$ for all $E \in \mathcal{N}$.

From Rudin

If $(X, \mathcal{M})$ is a measurable space, $(Y, \tau)$ is a topological space, and $f$ is a mapping of $X$ into $Y,$ then $f$ is said to be measurable provided that $f^{-1}(V) \in \mathcal{M}$ for every $V \in \tau.$

Finally from my first- year graduate course in analysis, in which we define measurable spaces and measurability in terms of $\sigma$-rings as opposed to $\sigma$-algebras, we have this proposition.

Let $(X, \mathcal{S})$ be a measurable space, $B$ a Banach space, and $f$ a function from $X$ to $B.$ Then $f$ is $\mathcal{S}$-measurable if and only if
1. $f(X)$ is a separable subset of $B,$ and
2. $f^{-1}(U) \cap C(f) \in \mathcal{S}$ for every open ball, $U,$ in $B$

where $C(f) = \{ x \in X : f(x) \neq 0 \}.$

The proposition above seems almost equivalent to Rudin's definition if we restrict it to $\sigma$-algebras since $$f^{-1}(U) \cap C(f) = f^{-1}(U) \cap f^{-1}(B \setminus \{0\}) = f^{-1}(U \setminus \{0\})$$ and singletons are closed in $B.$ But does the requirement that $f(X)$ be separable change the class of measurable functions if we let $(Y, \tau) = B$ with its topology induced by the norm?

Furthermore, Rudin's definition agrees with Folland's when $\mathcal{N}$ is the collection of Borel subsets, and we can always generate a $\sigma$-algebra from a topology. However, are there $\sigma$-algebras that cannot be generated by a topology? If so then it would seem that Folland's definition is "more general", otherwise they would be equivalent wouldn't they?

Fady Nakhla
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1 Answers1

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(In my opinion) Folland's definition is the "real" one. You are correct that Rudin's definition is secretly a special case of Folland's, if you put the Borel $\sigma$-algebra on $Y$ (that is, the fact that the inverse image of every open set is measurable implies that the inverse image of every Borel set is measurable).

Your third quote, with $\mathcal{S}$-measurability, looks closer to what I would call "strong measurability", which arises in the theory of vector-valued integration. See also: https://en.wikipedia.org/wiki/Bochner_measurable_function

pseudocydonia
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  • Yes, the definition we gave of measurability was quite similar to the one given in that article with a few differences. The simple S-measurable functions were required to have finite values instead of countable values, and the preimage of each singleton except zero were required to be measurable. Restricting the range to a finite set shouldn't change the definition much since we are taking limits correct? and the exception for {0} is what allows us to use sigma rings instead of algebras? Otherwise we would necessarily have that the whole set is a member of S. – Fady Nakhla Jul 26 '19 at 20:37