Local properties of Baire 1 functions

Question

A Baire 1 function $f:[0,1]\rightarrow \mathbb{R}$ need not be bounded. However, thanks to the Baire category theorem, we know there is $N\in \mathbb{N}$ and a sub-interval $(c, d) \subset [0,1]$ such that $(\forall x\in (c, d))( |f(x)|\leq N)$. In other words, a Baire 1 function may not be bounded on the unit interval, we know it is bounded on some sub-interval.

Similarly, a Baire 1 function $f:[0,1]\rightarrow \mathbb{R}^+$ may have infimum equal to $0$, but there is always some sub-interval $(c, d)\subset [0,1]$ where $f$ has non-zero infimum.

I am curious what other nice properties Baire 1 functions can have that fail on the unit interval, but do hold on some sub-interval. I would also be happy with examples for e.g. upper semi-continuous functions.

Answer 1

Let $ s:[0;1]\to[0;1]$ be the average binary digit partial function defined as follows:

$$ s(x)\ :=\ \lim_{n=\infty}\quad s_n(x) $$

where $$ s_n(x)\, :=\,\ \frac 1n\cdot \sum_{k=1}^n b_{-k}(x) $$ and $\ b_k(x)\ $ are the binary digits of $\ x.$

Borel's theorem tells us that $\ s(x)=\frac12\ $ for all $\ x\in[0;1]\ $ except for Lebesgue measure $\ 0.$ Nevertheless, the following popular question was open for a rather long time:

$\qquad$ does there exist $\ x\in(0;1)\ $ (an open interval)

$\qquad\qquad\qquad$ such that $\ s(x)=x\,?$

I answered this question and more in the year 1959; in particular, I replaced function $\ f(x):=x\ $ by any function from a larger set of functions:

Theorem   Let $\ f:[0;1]\to[0;1]\ $ be an arbitrary Baire 1 function. Then for arbitrary $ \emptyset \ne [a;b]\subseteq [0;1] $ there exist continuum many $\ x\in[a;b]\ $ such that $$ s(x)\ =\ f(x). $$