The Baire's simple limit theorem states that if the functions $f_n : \mathbb{R} \to \mathbb{R}$ are continuous and converge everywhere to a function $f$ then $f$ has a dense set of continuity points. We say that $f$ is of Baire class one when $f$ is a limit everywhere of continuous functions.

In the following, "measure" will refer to the Lebesgue measure.

**Question.** Assume that there is continuous functions $f_n : [0,1] \to \mathbb{R}$ and a measurable function $f$ such that $f(t)=\lim _{n \to \infty} f_n(t)$ for $t \in A \subset [0,1]$, a measurable set of measure $1$. Is it true that $f$ is equal almost everywhere to a function of Baire class $1$ ?

**N.B.** Yesterday I posted a trivial (and useless) version of the question and a user proposed $f_n=0$ and $f=1_{\mathbb{Q}}$, but $f=0$ a.e which does not provides a counter-example with this formulation.

The problem is that when we take continuous functions $f_n : [0,1] \to \mathbb{R}$ and a $f$ such that $f(t)=\lim _{n \to \infty} f_n(t)$ for $t \in A$ (of measure $1$) we are not sure that $(f_n(t))_{n \ge 0}$ converges and not even sure if $f_n(t)$ is defined.

**Example.** If $N_k=2^{q_k}$, $q_1=1$ and $q_{k+1}=2^{q_k}+q_k+1$. Setting $e_N(\theta)=e^{2i \pi \theta}$ for $\theta \in [0,1]$ one can see that : $$h:= \sum_{k \ge 1} \frac{e_{N_k}+e_{-N_k}}{k} \in L^2 \subset L^1$$
Thus there exists an increasing sequence $K_j \to \infty$ satisfying : $$h(\theta)= \lim_{j \to \infty} \sum_{k=1}^{K_j} \frac{2 \cos (2\pi K_j\theta)}{k} \text{ a.e }$$

Let $f_j : x \mapsto \sum_{k=1}^{K_j} \frac{2 \cos (2\pi K_j\theta)}{k}$ which is a continuous function. What my question ask here is to state whether $\lim f_j$ agrees almost everywhere with a function of Baire class $1$ i.e a function which is a simple limit of continuous functions. It is not far since the $f_j$ are continuous and that $(f_j(t))_{j}$ converges for a.e $t$, but NOT everywhere since for example $f_j(0)$ diverges.

If it is not clear please tell me what is unclear and I will clarify !

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