Baire Function may not be Continuous

Theorem
Let $X \subseteq \R$.

Let $f : X \to \R$ be a Baire function.

Then $f$ is not necessarily continuous.

Proof
Let:
 * $X = \closedint 0 1$

For each $n$, define the function $f_n : \closedint 0 1 \to \R$ to have:
 * $\map {f_n} x = x^n$

for each $x \in \closedint 0 1$.

Note that for $0 \le x < 1$, we have:
 * $x^n \to 0$

Note also that:
 * $\map {f_n} 1 = 1$

so:
 * $\map {f_n} 1 \to 1$

Define $f : \closedint 0 1 \to \R$ by:
 * $\ds \map f x = \begin{cases}0 & 0 \le x < 1 \\ 1 & x = 1\end{cases}$

for each $x \in \closedint 0 1$.

We have:
 * $\sequence {f_n}$ converges to $f$ pointwise.

Clearly $f$ is not continuous at $1$, since:
 * $\size {\map f x - \map f 1} = 1$

for any $x \in \hointr 0 1$.

So $f$ is a Baire function that is not continuous.

Also see

 * Continuous Real Function is Baire Function: this result shows that the converse of that result is false.
 * Set of Discontinuities of Baire Function is Meager: quantifies the discontinuity sets of Baire functions more precisely.