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.

  • 2
    $\begingroup$ 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. $\endgroup$ Commented Mar 12, 2023 at 15:07
  • $\begingroup$ @DaveL.Renfro Thanks for the advice. I have added the link for the reference in my post. $\endgroup$
    – user136524
    Commented Mar 13, 2023 at 5:19
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Mar 13, 2023 at 6:26

1 Answer 1


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 $$ \forall x,y: ~~ |x-y|< \min\{\delta(x),\delta(y)\} \Rightarrow |f(x)-f(y)| <\varepsilon $$

  • $\begingroup$ why do you consider $|x-x_0|<2$ and $|x-x_0|\geq 2$? $\endgroup$
    – user136524
    Commented Mar 15, 2023 at 14:20
  • 1
    $\begingroup$ @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 $|x-x_0| \ge 2$. $\endgroup$
    – Gerd
    Commented Mar 15, 2023 at 18:24

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