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Consider the unit interval $I=[0, 1]$ and assume that the function $f\colon I\to \mathbb R$ satisfies $$ f(t)=\lim_{n\to \infty} f_n(t), \qquad \text{for all }t\in I $$ where $$ f_n(t)=\lim_{j\to \infty} f_{n\, j}(t), \qquad \text{for all }t\in I $$ and $f_{n\, j}\in C(I)$.

Question. Is it true that there exists a sequence $g_n\in C(I)$ such that $$f(t)=\lim_{n\to \infty} g_n(t), \qquad \text{for all }t\in I?$$

The answer is positive if every "$\text{for all}$" is replaced by "$\text{for almost all}$", and the proof uses in an essential way Egorov's theorem to obtain some uniformity. (See Proposition 1.4.3 here). That's why I think that the answer to the present question is negative.

Giuseppe Negro
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2 Answers2

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The answer is negative. In fact if we define the Baire class $B_\alpha$ for countable ordinals $\alpha$ by saying $B_0=C(I)$, $B_{\alpha+1}$ is the set of pointwise limits of sequences in $B_\alpha$, and $B_\alpha=\bigcup_{\beta<\alpha}B_\beta$ for limit ordinals $\alpha$ then all the $B_\alpha$ are distinct.

Or so I've read; don't ask me to prove it. Your question asks whether $B_2=B_1$, and it's easy to give an example showing that that's not so.

Say $S=[0,1]\cap\Bbb Q$ and let $f=\chi_S$. It's clear that the characteristic function of any finite set is in $B_1$, and hence $f\in B_2$. It follows from the Baire Category Theorem that $f\notin B_1$:

If $(f_n)$ is a sequence of continuous functions then $\{x:\limsup f_n(x)>1/2\}$ is a $G_\delta$. But $S$ is not a $G_\delta$. Say $S=\{r_1,r_2,\dots\}$, and let $S_n=\{r_n,r_{n+1}\dots\}$. If $S$ were a $G_\delta$ then each $S_n$ would be a dense $G_\delta$ and hence $\emptyset$ would be dense.

David C. Ullrich
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  • This is a very clear answer but I must admit that I fail to see why ${\liminf f_n(x)>\frac12}$ is a $G_\delta$. I can rewrite that set as $$\displaystyle \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty \left{f_k(x)>\frac12\right}.$$ This seems to be a countable union of $G_\delta$ sets and I don't see the reason why it should be a $G_\delta$ set itself, could you please explain? – Giuseppe Negro May 26 '16 at 15:19
  • @GiuseppeNegro Oops. Change $\liminf$ to $\limsup$. Sorry... thanks. – David C. Ullrich May 26 '16 at 15:23
  • Great! Now it is $$\left{ x\ :\ \limsup_{n\to \infty} f_n(x) >\frac12\right} = \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty\left{ f_k(x)>\frac12\right}$$ which is a $G_\delta$. (Wrote this down for the eventual benefit of some reader). – Giuseppe Negro May 26 '16 at 15:49
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Ciao Giuseppe.

The answer is negative, as @DavidC.Ullrich has already pointed out. I can provide an explicit counterexample. Consider the function $q:\mathbb{Q}\backslash\{0\}\to\mathbb{N}$ associating to $x$ the unique natural number $q=q(x)$ such that we can write $x=\tfrac{p}{q}$ an irreducible fraction. Then define $$f(x) = \cases{q(x)&if $x\in\mathbb{Q}\backslash\{0\}$,\\0&else.}$$ This function is not the pointwise limit of any sequence of continuous functions (see e.g. this answer to a question about pointwise limits of continuous functions). Now it is easy to construct sequences as in your question. let $\{x_k\}_k$ be an enumeration of all non-zero rationals in $I$. For any number $x\in I$, let $\{g_j^{x}\}_j$ be a sequence of continuous functions converging pointwise to $\chi_{\{x\}}$ (easily constructed). Now take $$f_{n,j} = \sum_{k=1}^nq(x_k)g_j^{x_k}.$$ Then $$f_{n,j}\stackrel{j\to\infty}{\longrightarrow}\sum_{k=1}^nq(x_k)\chi_{\{x_k\}}=:f_n$$ pointwise, and $$f_n\stackrel{n\to\infty}{\longrightarrow}f$$ pointwise.

Community
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Daniel Robert-Nicoud
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  • Isn't this just a more complicated version of the example I gave? – David C. Ullrich May 25 '16 at 15:27
  • More or less, but it is more "explicit" in the sense that the link I provided gives a proof (not even relying on Baire's theorem) of the fact that my function $f$ is not a pointwise limit of continuous functions. Also, I gave the sequence of functions. That said, I like your answer more than mine as it is more conceptual ;) – Daniel Robert-Nicoud May 25 '16 at 16:48
  • FYI, your example and the example by @David C. Ullrich are both discontinuous at every point, but in fact examples exist having continuity sets that are dense in the reals. Baire gave the first such example, the characteristic function of the set of left endpoints for the Cantor middle thirds set, on p. 50 of his 1899 Ph.D. dissertation. In fact, the classification of functions as Baire 1, Baire 2, etc. (Baire didn't take this into the transfinite in 1899, however) and many of their basic properties are due to him. – Dave L. Renfro May 26 '16 at 15:54
  • @DaveL.Renfro That's interesting. The proof I gave for the characteristic function of the rationals works here as well, applying BCT on the Cantor set instead of on $[0,1]$. – David C. Ullrich May 26 '16 at 16:08
  • @David C. Ullrich: There's a theorem which I'm sure you know that says every Baire 1 function is continuous almost everywhere in the sense of Baire category (i.e. discontinuous at most on a first category set). It's natural to think the converse holds, but there's Baire's counterexample. A stronger result, whose converse does happen to hold, is that a Baire 1 function has the property that its restriction to any perfect set is continuous Baire-almost everywhere relative to that perfect set. Proving the stronger version is about the same as you can see, but the converse is a bit more involved. – Dave L. Renfro May 26 '16 at 17:22