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I am confused about the notion of Baire functions (real or complex valued) on a compact space $X$.

The set of Borel functions on $X$, $Bo(X)$ is defined to be the set of those functions $f$ for which $f^{-1}(U)$ is a Borel set, when $U$ is open.

On the other hand, the Baire functions of class $\alpha$, $Ba_{\alpha}(X)$, where $\alpha$ is a ordinal less than $\omega_1$, are defined iteratively as the pointwise limit of functions in the previous Baire classes, with $Ba_{0}(X)=C(X)$, continuous functions on $X$ (see here). Then the set of all Baire functions is defined to be $Ba(X):=\cup_{\alpha<\omega_1}Ba_\alpha$

Clearly, $Bo(X)\supset Ba(X)$. Now, this article assumes my definition of Baire functions, and this article (see "Comparison with Baire functions"), says that the Baire functions are the smallest set of those real functions closed under pointwise limits and containing continuous functions.

This however, seems to imply that $Ba(X)$ is closed under pointwise limits.

Is this obvious? I cannot seem to show this for $Ba(X)$.

(If we restrict ourselves to bounded functions, is the statement true for uniform limits of Baire functions?)

Any help is much appreciated!

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  • $\begingroup$ In general $\lim_n \lim_m f_{nm}$ with each $f_{nm}$ continuous cannot be expressed as a limit of a single sequence of continuous functions. It is not true that pointwise limit of $B_a(X)$ is closed under pointwise limits. $\endgroup$ – Kavi Rama Murthy Oct 19 '18 at 23:23
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It's quite easy, really: let $f_n$ be a sequence of Baire functions (so from $\operatorname{Ba}(X)$) tending pointwise to $f$. By definition for each $n$, $f_n \in \operatorname{Ba}_{\alpha_n}(X)$ for some $\alpha_n < \omega_1$. In $\omega_1$, every countable set has an upper bound so there is some $\beta_0 < \omega_1$ such that for all $n$ we have $\alpha_n < \beta_0$ and so, as

$$\forall \gamma < \delta < \omega_1: \operatorname{Ba}_\gamma(X) \subseteq \operatorname{Ba}_\delta(X)$$

(the $\operatorname{Ba}_\alpha(X)$ are easily seen to be increasing in $\alpha$) we know that all $f_n \in \operatorname{Ba}_{\beta_0}(X)$. So by definition $f \in \operatorname{Ba}_{\beta_0 + 1}(X)$ as a pointwise limit of functions from the previous level (or in your definition, it's already in $\operatorname{Ba}_{\beta_0}(X)$). And so $f \in \operatorname{Ba}(X)$.

Now you maybe understand better why we can stop at stage $\omega_1$ to get "countable closure properties".. The Borel hierarchy works the same way.

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  • $\begingroup$ Thanks very much - I was stuck on why there would be an upper bound for the countable family, but I guess my knowledge of ordinals is incomplete (I didn't realize there would be $\beta_0<\omega_1$ upperbound). $\endgroup$ – Merry Oct 22 '18 at 21:44
  • $\begingroup$ @Merry $\omega_1$ is the first uncountable ordinal. A countable set of countable ordinals has its union, which is also an ordinal and still countable, and so $< \omega_1$ by definition. $\endgroup$ – Henno Brandsma Oct 22 '18 at 21:46
  • $\begingroup$ That makes sense from a set theoretic perspective! Thanks. $\endgroup$ – Merry Oct 22 '18 at 21:58
  • $\begingroup$ @Merry descriptive set theory is a nice mix of set theory and topology. $\endgroup$ – Henno Brandsma Oct 22 '18 at 22:00
  • $\begingroup$ Indeed! It's something I have only recently encountered. $\endgroup$ – Merry Oct 23 '18 at 2:02

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