Does pointwise convergence imply uniform convergence on a large subset?

Question

Suppose $f_n$ is a sequence of real valued functions on $[0,1]$ which converges pointwise to zero.

1. Is there an uncountable subset $A$ of $[0,1]$ so that $f_n$ converges uniformly on $A$?

2. Is there a subset $A$ of $[0,1]$ of cardinality the continuum so that $f_n$ converges uniformly on $A$?

Background: Egoroff's theorem implies that the answer to (2) is yes if all $f_n$ are Lebesgue measurable. It is not hard to show that the answer to (1) is yes if you change "uncountable" to "infinite".

Motivation: I thought about this question while teaching real analysis this term but could not solve it even after looking at some books, googling, and asking some colleagues who are much smarter than I, so I assigned it as a problem (well, an extra credit problem) to my class. Unfortunately, no one gave me a solution.

ADDED 11-12-10: Thanks for all the great answers. I accepted Jonas' answer since it was the first one.

Answer 1

I did some Googling and came up with something that looks relevant, Theorem 10 quoted below from Morgan's Point set theory. It cites works of Sierpiński from the late 1930s, but I can't tell what works are cited because the preview won't let me see that page in the references.

The existence of a linear set having the power of the continuum that is concentrated on a denumerable set is equivalent to the existence of a pointwise convergent sequence of functions of a real variable that does not converge uniformly on any uncountable set.

Answer 2

A simple diagonalization argument gives a counterexample if you assume the continuum hypothesis. Let a modulus of convergence be a sequence $\delta:{\mathbb N}\to(0,\infty)$ which converges to 0, and say that $\delta>\delta'$ if $\delta(n)>\delta'(n)$ for sufficiently large $n$. By diagonalizing, for any countable set $\{\delta_n\}$ of sequences there is some $\Delta$ (which still converges to 0) such that $\Delta>\delta_n$ for all $n$.

We say that a sequence $x_n$ converges to $x$ with modulus $\delta$ if $|x_n-x|<\delta(n)$ for all $n$. We then have that $f_n$ converges to $f$ uniformly iff there is a single $\delta$ such that $f_n(x)$ converges to $f(x)$ with modulus $\delta$ for all $x$.

Now assuming the continuum hypothesis, the set of all moduli and the set of all points in $[0,1]$ are both in bijection with $\omega_1$; let $(\delta_\alpha)_{\alpha<\omega_1}$ and $(x_\alpha)_{\alpha<\omega_1}$, respectively, be enumerations. For each $\alpha$, let $\Delta_\alpha$ be a modulus such that $\Delta_\alpha>\delta_\beta$ for all $\beta<\alpha$ (since there are only countably many such $\beta$). Now let $f=0$ and define $f_n(x_\alpha)=\Delta_\alpha(n)$. Then $f_n(x_\alpha)$ converges to $f(x_\alpha)=0$ for all $x_\alpha$. However, for any fixed modulus $\delta=\delta_\beta$, $f_n(x_\alpha)$ cannot converge with modulus $\delta$ for $\alpha>\beta$ by construction, so there are only countably many points on which $f_n$ converges uniformly with modulus $\delta$.

Answer 3

To add a bit to Jonas Meyer's answer.

Sierpiński's result was first published in C.R. Soc. Sc. Varsovie 1928, p. 84-87. It was reproduced in his monograph "Hypothèse du continu" (Lwów, 1934, p. 52):

Proposition $C_9$. Il existe une suite infinie convergente de fonctions d'une variable réelle
$f_1(x)$, $f_2(x)$, $f_3(x),...$ qui convergent non uniformément sur tout ensemble indénombrable.

Sierpiński effectively derived it from the statement which is implied by the continuum hypothesis (Ibid. p. 36):

Proposition $C_1$. Il existe un ensemble linéaire $N$ de puissance du continu qui admet un ensemble au plus dénombrable de points communs avec tout ensemble (linéaire) parfait non-dense.

Answer 4

This paper by R. Pinciroli claims that the stronger statement:

$f_n$ converges uniformly on sets of outer measure arbitrarily close to 1

is undecidable in ZFC.

Answer 5

Also, Shinoda (1973) proved that if Martin's Axiom and $\neg CH$ hold, then the answer to 1) si affirmative.