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. $\endgroup$ – Mateusz Wasilewski Nov 15 '16 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. $\endgroup$ – Jacques Mardot Nov 15 '16 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. $\endgroup$ – Mateusz Wasilewski Nov 15 '16 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. $\endgroup$ – Jacques Mardot Nov 15 '16 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. $\endgroup$ – Jacques Mardot Nov 15 '16 at 15:49


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.

| cite | improve this answer | |
  • $\begingroup$ 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 ... $\endgroup$ – Jacques Mardot Nov 15 '16 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.