# Pointwise convergence imples uniform convergence in an infinite subset

I came upon this statement in a stack answer.

Statement : If $$f_n$$ is a sequence of real valued functions (not necessarily continuous or measurable) on $$[0,1]$$ such that $$f_n$$ converges point-wise to $$0$$, then there exists an infinite subset of $$[0,1]$$ where the convergence is uniform.

I couldn't prove it. I believe the claim is true because $$[0,1]$$ is uncountable but the set of sequences is countable only.

Any help would be appreciated in assistance to how to prove it.

• If $\mathfrak{c}$ is a real-valued measurable cardinal then this follows from Egorov's theorem... Jul 12, 2021 at 19:25
• mathoverflow.net/questions/45784/… says this is "not hard" but doesn't give a proof. (It mainly discusses whether "infinite" can be strengthened to "uncountable".) Jul 12, 2021 at 19:27
• Crossposted at math.stackexchange.com/questions/4196015/…: when you do this, please make sure each question contains a link to the other. Jul 12, 2021 at 19:37
• @NateEldredge apologies. I'll remember that in the future. Jul 12, 2021 at 19:39
• @NateEldredge I saw that answer. I have commented on the question seeking an explanation, but the question is 10yrs old, so don't know how helpful that would be. Jul 12, 2021 at 19:43

For every sequence $$(F_n)_{n \in \omega}$$ with $$F_n:[0,1] \rightarrow \mathbb{R}$$ converging pointwise to $$0$$, we can associate to every $$x \in [0,1]$$ an $$f_x \in \mathbb{\omega}^\omega$$ in the following way: Set $$f_x(m):= \min\{n \in \mathbb{\omega}\,\, \colon \,\, \forall n' \geq n \,\,\,\, \vert F_{n'}(x) \vert < \frac{1}{m}\}$$.

Now the proof works in two steps.

In the first step we show that there exists $$f^* \in \omega^\omega$$ such that for every $$k \in \omega$$ the set $$\{x \in [0,1] \,\, \colon \,\, f_x \restriction k \leq f^* \restriction k\}$$ is uncountable.

We will construct such an $$f^* \in \omega^\omega$$ by induction on $$k \in \omega$$: Assume that $$f^* \restriction k$$ has already been constructed, and we have that the set $$\{x \in [0,1] \,\, \colon \,\, f_x \restriction k \leq f^* \restriction k\}$$ is uncountable. Since $$\{x \in [0,1] \,\, \colon \,\, f_x \restriction k \leq f^* \restriction k\}= \bigcup_{l \in \omega}\,\, \{x \in [0,1] \,\, \colon \,\, f_x \restriction (k+1) \leq (f^* \restriction k)^\frown l\}$$ we find $$l \in \omega$$ such that $$\{x \in [0,1] \,\, \colon \,\, f_x \restriction (k+1) \leq (f^* \restriction k)^\frown l\}$$ is uncountable. Now set $$f^*(k)=l$$ and we see that $$f^*\restriction (k+1)$$ has the required properties.

In the second step we inductively construct $$(x_k)_{k \in \omega} \subseteq [0,1]$$ injective and $$(f_k)_{k \in \omega} \subseteq \omega^\omega$$ increasing such that for every $$k \in \omega$$ we have $$f^* \leq f_k$$ , $$f_{x_k} \leq f_k$$ and $$f_k \restriction (k+1) = f_{k+1} \restriction (k+1)$$. Once we have shown this, we can define $$g \in \omega^\omega$$ such that $$g(k):=f_k(k)$$ and see that $$f_{x_k} \leq g$$ for every $$k \in \omega$$. But this proves that $$(F_n)_{n \in \omega}$$ converges uniformly on $$(x_k)_{k \in \omega}$$.

To this end assume that $$x_0,...,x_{k-1}$$ and $$f_0,...f_{k-1}$$ with the required properties have already been constructed. Since $$\{x \in [0,1] \,\, \colon \,\, f_x \restriction k \leq f^* \restriction k\}$$ is uncountable, we can find $$x_k \in \{x \in [0,1] \,\, \colon \,\, f_x \restriction k \leq f^* \restriction k\}$$ different from $$x_0,...,x_{k-1}$$. Set $$f_k(m):=\max\{f_{k-1}(m), f_{x_k}(m)\}$$, and note that since $$f_{x_k} \restriction k \leq f^* \restriction k \leq f_{k-1} \restriction k$$, we have $$f_{k-1}\restriction k =f_k \restriction k$$. This finishes the proof.

• Here $\omega$ denotes the set of natural numbers, and for every $k \in \omega$ we have $k=\{0,...,k-1\}$. Also note that the proof cannot be much simpler, since your statement holds iff for every $(f_x)_{x \in [0,1]} \subseteq \omega^\omega$ there exists $X \subseteq [0,1]$ infinite and $g \in \omega^\omega$ such that $f_x \leq g$ for every $x \in X$. Jul 12, 2021 at 23:38
• Thanks for the answer. Jul 13, 2021 at 5:05