# Baire one function and its discontinuity point

Let $$f:\mathbb{R} \to \mathbb{R}$$ be a function which is defined by \begin{align*} f(x)=\begin{cases} 0,\quad x\leq 2,\\ 1,\quad x>2. \end{cases} \end{align*} The function $$f$$ above is Baire one function. Recently, I found the definition of Baire one function at a point using $$\epsilon-\delta$$ notation in Theorem 3.10 and Theorem 3.11 from this article , I write it below by using a function from $$\mathbb{R}$$ to $$\mathbb{R}$$.

A function $$f$$ is Baire one at $$x_0\in\mathbb{R}$$ if for any $$\epsilon>0$$, there exists a positive function $$\delta:\mathbb{R}\to\mathbb{R}$$, such that for any $$x\in \mathbb{R},$$ $$|x-x_0|<\min\{\delta(x),\delta(x_0)\} \Rightarrow |f(x)-f(x_0)|<\epsilon.$$

Since $$f$$ is continuous at every point $$x_0\neq 2$$, then it is true that $$f$$ is Baire one at every point $$x_0\neq 2$$.
Now, I am curious, how about if $$x_0=2$$. Is the function $$f$$ Baire one at $$x_0=2$$?
I think $$f$$ is not Baire one at $$x_0=2$$. But I do not know how prove it properly.
Thanks for any help.

• You should always give a specific reference when asking about the statement of something not particularly well known. In this case, a similar notion appears to have arisen in Lee/Tang/Zhao 2001 and has since appeared in a few papers, such as Fenecios/Cabral 2013 and Balcerzak/Karlova/Szuca. Examination of these papers suggests that you may have misstated something. Commented Mar 12, 2023 at 15:07
• @DaveL.Renfro Thanks for the advice. I have added the link for the reference in my post. Commented Mar 13, 2023 at 5:19
• Thanks for the link. Others will have to investigate this because the paper is behind a paywall and I do not have access, and I do not plan on making a trip to a university library in the near future. Commented Mar 13, 2023 at 6:26

The $$\varepsilon$$-$$\delta$$ characterization of Baire one functions is not a pointwise condition: Fix some $$x_0$$ and set $$\delta(x)=\frac{|x-x_0|}{2} ~ (x \not=x_0), \quad \delta(x_0)=1.$$
Assume that some $$x \not= x_0$$ satisfies $$|x-x_0|< \min\{\delta(x),\delta(x_0)\}= \min\{|x-x_0|/2,1\}$$.

1. If $$|x-x_0| < 2$$, then $$|x-x_0| < |x-x_0|/2$$, a contradiction.
2. If $$|x-x_0| \ge 2$$, then $$|x-x_0| < 1 < 2 \le |x-x_0|$$, a contradiction.

Thus $$|x-x_0|< \min\{\delta(x),\delta(x_0)\}$$ implies $$x=x_0$$. Then $$|f(x)-f(x_0)|= 0 <\varepsilon$$. So each function $$f$$ has this property at each point $$x_0$$.

The $$\varepsilon$$-$$\delta$$ characterization of Baire one functions reads (see the first link from the comment of Dave L. Renfro):

For any $$\varepsilon > 0$$ there is a function $$\delta:\mathbb{R} \to (0,\infty)$$ such that $$\forall x,y: ~~ |x-y|< \min\{\delta(x),\delta(y)\} \Rightarrow |f(x)-f(y)| <\varepsilon$$

• why do you consider $|x-x_0|<2$ and $|x-x_0|\geq 2$? Commented Mar 15, 2023 at 14:20
• @user136524 Since $\min\{|x-x_0|/2,1\} = |x-x_0|/2$ if $|x-x_0|< 2$, and $\min\{|x-x_0|/2,1\} = 1$ if $|x-x_0| \ge 2$.
– Gerd
Commented Mar 15, 2023 at 18:24