Some Locally Uniform Convergence

Question

Let $(f_n)$ be a sequence of functions where $f_n:[0,1] \to \mathbb{R}$ for all $n \in \mathbb{R}$. If $\lim f_n(x) = 0$ for every $x \in [0,1]$, prove that there exist $[a,b] \subseteq [0,1]$ and a natural number $N_0$ such that for all $x \in [a,b]$ and $n \geq N_0$, $|f_n(x)| < 1$ holds.

I wonder if this question is correct. My idea was to choose a point, then try to make a small interval centered at that point, but I noticed that the idea required continuity (the question doesn't specifically say continuous functions). So now I'm stuck to find the correct approach and wondering whether the question is correct at all.

Answer 1

This is false without continuity. Let $(r_n)$ be an ennumeration of rationals in $[0,1]$. Let $f_n(x) =0$ for $x \neq r_n$ and $f_n(r_n)=1$. Then $f_n(x) \to 0$ for each $x$. If the conclusion holds then $[a,b]$ would contain only a finite number of rational numbers, a contradiction.

With continuity this is an easy application of Baire Category Theorem.

Answer 2

Elaborating on the Baire category argument suggested by the other answerer, suppose each $f_n$ is continuous. Let $U_n = \{x \in [0,1] : f_j(x) \leq 1/2\text{ for all }j\geq n\}$. Then each $U_n$ is closed, and $$[0,1] = \bigcup_{n\geq 1} U_n,$$ so the completeness of $[0,1]$ means that there is a $U_n$ with nonempty interior. In particular, there is a $U_n \subset [0,1]$ and an interval $(a,b) \subset [a,b] \subset U_n$. Then $|f_j(x)| \leq 1/2 < 1$ for each $j\geq n$ and $x\in [a,b]$.