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 !

  • $\begingroup$ For any $L^1$-function, in particular any bounded function, there exists a sequence of continuous functions that converges to it almost everywhere; continuous functions are dense in $L^1$ and you get convergence almost everywhere by passing to a subsequence, using the Borel-Cantelli lemma. $\endgroup$ Commented Nov 15, 2016 at 9:10
  • $\begingroup$ I understand but your sequence $(f_n)$ does not converge at every point ? The convergence everywhere of $(f_n)$ is really required. $\endgroup$
    – user94415
    Commented Nov 15, 2016 at 10:42
  • $\begingroup$ If $(f_n)$ converges everywhere to some $g$ then $f=g$ almost everywhere, so $f$ equals almost everywhere to a function of Baire class $1$. I'm not sure if I understand the question properly. $\endgroup$ Commented Nov 15, 2016 at 10:50
  • $\begingroup$ I will try again : let say we have a function $f \in L^1$ and continuous functions $f_n:[0,1] \to \mathbb{R}$ such that $f=\lim f_n$ on a set $A$ of measure $1$. So $f$ is a.e equal to $g=\lim f_n$ but a priori $g$ is not of Baire class one because the convergence $\lim f_n(t)$ holds only for $t \in A$ and not everywhere. $\endgroup$
    – user94415
    Commented Nov 15, 2016 at 12:02
  • 2
    $\begingroup$ Yes, the sequence $(f_n)$ is required to converge everywhere toward a function $g$ such that $g=f$ a.e. The second comment of Mateusz does not answer the question since he does not provide a construction of such $(f_n)$ converging everywhere. The first comment does only deals with $(f_n)$ converging almost everywhere. $\endgroup$
    – user94415
    Commented Nov 15, 2016 at 15:49

1 Answer 1



Your first hypothesis is true for any Lebesgue measurable function $f$; there is a sequence of continuous functions $f_n$ such that $f_n \to f$ almost everywhere. This is standard.

But there do exist Lebesgue measurable (even Borel) functions which are not a.e. equal to any function of Baire class 1.

For instance, it's a fairly standard exercise to construct a Borel set $B \subset [0,1]$ such that for any nontrivial interval $I$, we have $0 < m(B \cap I) < m(I)$. See this Math.SE post for details. Take $f = 1_B$.

Suppose $g = f$ a.e.; let $C_1 = g^{-1}(\{1\})$ and $C_0 = g^{-1}(\{0\})$, so $m(C_1 \triangle B) = 0$ and $m(C_0 \triangle B^c) =0$. As such, for any interval $I$ we have $m(C_i \cap I) > 0$. So in particular, on any interval, there are points at which $g=0$ and points at which $g=1$. Hence $g$ is nowhere continuous and thus not Baire class 1.

See also https://math.stackexchange.com/questions/15088/is-every-lebesgue-measurable-function-on-mathbbr-the-pointwise-limit-of-con. It's mentioned there that every Lebesgue measurable function is a.e. equal to some function of Baire class 2.

  • $\begingroup$ Fine, perfectly what I wanted ! I expected the result to be true (sadly it is not). And for my example do you think that $h$ is a.e equal to a function of Baire class $1$ ? I think $h$ has at least one point of continuity but showing it seems really challenging ... $\endgroup$
    – user94415
    Commented Nov 15, 2016 at 17:02

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