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(Big) Rudin's "Real and Complex Analysis" defines (definition 1.3) a measurable function from a measurable space into a topological space as one that has the property that the inverse image of every open set in the range space is measurable in the domain space.

Is this definition somehow more general than the one between two measurable spaces that pulls back measurable sets to measurable sets? My understanding is that topologies and sigma-algebras do not necessarily coincide, so I'm not sure why Rudin is using this definition for measurable functions.

Peter
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3 Answers3

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This definition isn't any more or less general, it's just the way to define a measurable function from a measurable space into a topological space. Until you've given it a topology, a measurable space is just that, and vice versa - until you define a sigma algebra, a topological space is not a measurable space. Rudin uses this definition because he needs topological structure (and not measurable structure) on his target space.

Also, it's always good practice to go through each theorem and figure out which structures were necessary in the proof.

icurays1
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  • "Rudin uses this definition because he needs topological structure (and not measurable structure) on his target space." Could you explain what could be reasons for that? I believe most current texts use measurable structures on both the source and the target, which seems more natural and convenient, and leaves a possibility of plugging-in a topology-induced (i.e. Borel) sigma-algebra. Ain't it so? – paperskilltrees Nov 11 '22 at 18:12
  • My question "Why does Rudin do that?" is also asked and addressed here: Rudin's definition on measurable function Yet maybe there are other reasons than simply working in $\mathbb{R},\mathbb{C}$, which have a standard topology? – paperskilltrees Nov 11 '22 at 18:17
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What Rudin is defining is usually called Borel measurability. If the range has a topology $\tau$, then it's natural to consider the $\sigma$-algebra $\sigma(\tau)$ generated by the topology, called the Borel $\sigma$-algebra.

General definition for measurability for maps between measurable spaces is as you said.

Notice that to check Borel-measurability, it is sufficient to check the generators $\tau$ by minimality of $\sigma(\tau)$.

The point of working with the Borel $\sigma$-algebra is to have compatibility between the two structures on your set. One example of a result displaying the link between $\tau$ and $\sigma(\tau)$ is the Riesz representation theorem:

Theorem Let $(X, \tau)$ be compact Hausdorff, every positive linear functional is integration with respect to some positive regular measure $\mu$ on $\sigma(\tau)$.

A measure is regular if measures of a measurable set can be approximated by open sets from above and compact sets from below. Notice openness and compactness are topological properties. So the theorem says integration against measures compatible with the topology coincide with linear functionals on $C(X)$.

Daniel Fischer
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Michael
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I encounter this different definition of measurable function in Rundin's book from the general/common definition in other Real Analysis/measure books.

My understanding is that in Real/complex space, Topological structure is embedded in Measurable space, because every infinite union sequence is countable. that is, if it is the measurable space in range, it is also a topological space.

JamesWang
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  • First, your response addresses why one would use a sigma-algebra instead of a topology, while the question is more why one would use a topology instead of a sigma-algebra. Second, if I understand correctly, your reasoning shows (at most) that a sigma-algebra contains a topology. It does not show that there is a topology inducing a given sigma-algebra. Neither does it show that such a topology is unique. Indeed, non-uniqueness is discussed here: Does the Borel-σ -Algebra define the topology of a Set? – paperskilltrees Nov 11 '22 at 18:33