Recently I read this post, which gives examples of Baire class 2 functions. I have also been reading the Wikipedia article on Baire functions. The article claims that "Henri Lebesgue proved that ... there exist functions that are not in any Baire class." I would like to see an example of such a function.

I am having a hard time finding anything that directly addresses my specific question. What I know so far is the definition: such a function is not the pointwise limit of any sequence of Baire class $\alpha$ functions, for any countable ordinal number $\alpha$. I have also found this article, which shows that any such function is not Lebesgue measurable.

I would also like to see some details of Lebesgue's proof that such functions exist. To be more specific, I would like to know whether it is constructive or is only an existence result.


2 Answers 2


I believe Lebesgue's proof was nonconstructive, and went as follows. There are at most continuum many countable ordinals, and only continuum many functions in each Baire class (exercise - how many countable subsets does a set of size continuum have?), so only continuum many Baire functions total. But there are two to the continuum many functions. (Note the similarity with the proof that there is a non-Borel set of reals, even a measurable one!)

Meanwhile, I don't think a constructive proof can truly exist, since it is consistent with ZF that every function is Baire. This is the case in models where the reals are a countable union of countable sets.

That said, here is a sorta-example: biject the Baire functions with the reals and diagonalize. That is, let $f $ be a subjection from the reals to the Baire functions, and let $g (r)=f (r)(r)+1. $


The Baire functions (starting with $\mathcal{C}_{00}(\mathbb{R})$ or $\mathcal{C}_b(\mathbb{R})$), or rather, the union of all Baire classes is the family of Borel measurable functions. Any Lebesgue-measurable function that is not Borel will do (The axiom of choice plays a role here). See for example Cohn. D., Measure Theory, 2nd edition, Birkhäuser, 2013, pp. 48 for a concrete example.

Here is a brief explanation of how the collection of all Baire functions is done thorough a transfinite inductive procedure:

Suppose $\mathcal{V}$ is a collection of real valued functions defined an a set $\Omega$. $\mathcal{V}$ is said to be sequentially closed if for any convergent sequence $(f_n:n\in\mathbb{N})\subset \mathcal{V}$, $f:=\lim_nf_n\in\mathcal{V}$. It is easy to check that the intersection of sequentially closed families is sequentially closed: suppose $\mathfrak{F}$ is a collection of sequentially closed families. If $(f_n:\in\mathbb{N})\subset\bigcap\mathfrak{F}$ and $f=\lim_nf_n$ exists, then for any $\mathcal{F}\in\mathfrak{F}$, $(f_n:n\in\mathbb{N})\subset\mathcal{F}$ and so, $f\in\mathcal{F}$.

Suppose $\mathcal{E}$ is a collection of real-valued functions defined on a set $\Omega$, let $\mathcal{E}^\sigma$ denote the smallest sequentially closed family that contains $\mathcal{E}$. This family exists from the observation above and since $\mathbb{R}^\Omega$ is clearly sequentially closed and contains $\mathcal{E}$.

Now, consider the following transfinite induction procedure: Let $\mathcal{E}_0=\mathcal{E}$; if $\mathcal{E}_\alpha$ has been defined for all ordinals $\alpha<\beta$, define $\mathcal{E}_\beta$ as \begin{align} \mathcal{E}_\beta=\left\{\begin{array}{lcr} \{f\in\mathbb{R}^\Omega: f=\lim_nf_n\,\text{for some}\,(f_n:n\in\mathbb{N})\subset\mathcal{E}_\alpha\} &\text{if}& \beta=\alpha+1\\ \bigcup_{\alpha<\beta}\mathcal{E}_\alpha &\text{if}& \beta\neq\alpha+1 \end{array} \right. \end{align} That is, if $\beta$ is the successor of $\alpha$, then $\mathcal{E}_\beta$ is the collection of pointwise limits of sequences in $\mathcal{E}_\alpha$, and if $\beta$ is not the successor of $\alpha$, then $\mathcal{E}_\beta=\bigcup_{\alpha<\beta}\mathcal{E}_\alpha$. It follows by construction that $\mathcal{E}_\beta\subset\mathcal{E}^\sigma$ for any ordinal $\beta$. The collection $\mathcal{E}_\beta$ is the $\mathcal{E}$-Baire familly of class $\beta$. (The case $\mathcal{E}=\mathcal{C}_b(\mathbb{R})$ yields the commonly known Baire classes).

It is easy to check that this construction stabilizes after the first uncountable ordinal $\omega_1$, for $\mathcal{E}_{\omega_1}$ is sequentially closed: suppose $(f_n:n\in\mathbb{N})\subset \mathcal{E}_{\omega_1}$, and $f_n\xrightarrow{n\rightarrow\infty} f$ pointwise to some function $f\in\mathbb{R}^{\Omega}$. Then there is a sequence of countable ordinals $\alpha_n$ such that $f_n\alpha_n$. As $\bigcup_n[0,\alpha_n]$ is countable and contained in $P_{\omega_1}:=[0,\omega_1)$, there is $\gamma\in P_{\omega_1}$ such that $\alpha_n\leq \gamma$ for all $n\in\mathbb{N}$; hence, $(f_n:n\in\mathbb{N})\subset \mathcal{E}_{\gamma}$ and so, $f\in \mathcal{E}_{\gamma+1}\subset\mathcal{E}_{\omega_1}$.

  • This statement appears (without proof) in Bichteler, K., Stochastic Integration with Jumps, Combridge U. Press, 2002., pp. 392., and no special structure on $\mathcal{E}$ (Stone lattice and/or ring for example) is assumed.

  • The special case $\mathcal{E}=C_b(\mathbb{R})$ for instance yields that the space $\mathscr{M}(\mathbb{R})$ of Borel measurable functions is the same as $(\mathcal{C}_b(\mathbb{R}))_{\omega_1}$. Indeed, clearly $\mathcal{E}^\sigma$ is contained in the space of all real-valued measurable functions $\mathscr{M}(\mathbb{R})$. Notice that the indicator function of any bounded open interval $I$ is in $\mathcal{E}_1$. The collection of sets in $\mathbb{R}$ whose indicator functions are in $\mathcal{E}_{\omega_1}$ is a $\sigma$-algebra; Hence $\mathscr{M}(\mathbb{R})\subset\mathcal{E}^\sigma$.


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