# Baire one function with a finite supremum on closed interval.

Let $f_n$ be a sequence of continuous functions on $\mathbf R$ that converge at every point. Prove there exist an interval and a number $M$ such that $\operatorname{sup}_n |f_n|$ is bounded by $M$ on that interval.

As a Baire one function on a compact subset of $\mathbf R$ in general need not even be bounded, I have no idea how to approach this problem, except through some trick using the Baire category theorem or something similar.

For any $M \in \mathbb{N}, k \in \mathbb{N}$, set $$G_{k,M} = \{x \in \mathbb{R} : |f_k(x)| \leq M\}$$ and $$F_M := \bigcap_{k=1}^{\infty} G_{k,M} = \{x \in \mathbb{R} : \sup_{k\in \mathbb{N}} |f_k(x)| \leq M\}$$ Then each $F_M$ is closed and by pointwise boundedness $$\mathbb{R} = \bigcup_{M=1}^{\infty} F_M$$ By Baire Category, $\exists M \in \mathbb{N}$ such that $F_M$ has non-empty interior. So $\exists$ an interval $J \subset F_M$ such that $$\sup_{k\in\mathbb{N}}|f_k(x)| \leq M \quad\forall x \in J$$