# Examples of Baire Class $\xi+1$ but not $\xi$ functions for each countable ordinal $\xi.$

We say that $f:\mathbb{R}\to\mathbb{R}$ is of Baire Class $1$ if it is a pointwise limit of a sequence of continuous functions.

One can generalize the definition above by taking pointwise limit of each 'previous' level(s) to obtain 'next' level. More precisely,

For any countable ordinal $\xi\geq 1,$ we say that $f:\mathbb{R}\to\mathbb{R}$ is of Baire Class $\xi$ if it is a pointwise limit of a sequence of Baire Class $\zeta$ functions where $\zeta<\xi.$

The generalization covers the case $\xi=1$ as every continuous function is of Baire Class $0.$

When $\xi = 2,$ it is well-known that $\chi_\mathbb{Q}$ is of Baire Class $2$ as it is a pointwise limit of $(g_n)_{n=1}^\infty$ where $g_n(x) = \max\{0,1-n d(x,K)\}$ and $K$ is a finite collection of rationals. (extracted from Wiki) Since $\chi_\mathbb{Q}$ is discontinuous everywhere, so it is not of Baire Class $1.$

This MSE post also contains other Baire Class $2$ functions.

However, I fail to obtain any Baire Class $3$ function and above.

Question: For each countable ordinal $\xi\geq 3,$ what are some examples of Baire Class $\xi+1$ but not $\xi$ function by using the pointwise limit definition?

• You might have a look, for example, at Section 16 in van Rooij, Schikhof: A Second Course on Real Functions. This part of the book is essentially about showing that $\mathscr B_{\xi+1} \setminus \mathscr B_\xi$ is none-empty for a countable ordinal $\xi$. – Martin Sleziak May 31 '18 at 10:28
• It is possible to explicitly construct Borel sets that are exactly at the level $\xi$ of the Borel hierarchy, but the construction is fairly unenlightening and involves repeated diagonalization over the previous levels: see here for the simplest account I was able to give. I don't think you should expect much better. – Gro-Tsen May 31 '18 at 13:55
• The buzzword here is "the Baire hierarchy does not collapse". – Nate Eldredge May 31 '18 at 14:30
• @Nate: Must have had a good architect! – Asaf Karagila May 31 '18 at 15:43
• See R. D. Mauldin's 1973 paper On Borel measures and Baire's class $3$ (proof on pp. 310-311) for a "naturally occurring" Baire $3$ function that is not Baire $2$, and see my comments in this 14 May 2009 sci.math post. – Dave L Renfro Jun 1 '18 at 8:29

A somewhat concrete example of function which is Baire class $\zeta$ but not Baire class $\gamma$ for any $\gamma < \zeta$ is the $\zeta$-th Turing jump. This is essentially the iterated version of Shoenfield's limit lemma. The (iterated) Turing jump naturally gives us a function $J : \{0,1\}^\omega \to \{0,1\}^\omega$, but of course we can embed $\{0,1\}^\omega$ into $\mathbb{R}$ as the Cantor middle set and extend the function, if we prefer to work with $\mathbb{R}$.
Another approach is to piggy-back on the non-collapse of the Borel hierarchy: By the Banach-Lebesgue-Hausdorff theorem, Baire class $\zeta$ and the $\Sigma^0_{1+\zeta}$-measurable functions coincide. So the characteristic function of a $\Sigma^0_{1+\zeta}$-complete set is Baire class $\zeta+1$, but not Baire class $\zeta$.
The example given in the question is of course a special case of the second procedure. A general construction of $\Sigma^0_{1+\zeta}$-complete sets is found eg in this answer