Prove that there exists a sequence of uniformly bounded measurable function such that ${f_n(x)}$ does not converge uniformly on any unbounded set

Question

Question: Prove that there exists a sequence $\{f_n(x)\}, x\in \mathbb{R}$ such that $\lim\limits_{n\to\infty} f_n(x)=f(x), \forall x\in \mathbb{R}$, $|{f_n(x)}|\leq M,\forall x\in \mathbb{R},\forall\mathbb{n}\in \mathbb{N}$ and $\{f_n(x)\}$ does not converge uniformly on any unbounded set $E$.

My effort: I was thinking about some common functions defined on $[0,1]$, such as $f_n(x)=x^n$ or $f_n(x)=\begin{cases}1-nx\quad x\in [0,\frac{1}{n}]\\ 0, x\in [\frac{1}{n},1]\end{cases}$ and try to extend them to $\mathbb{R}$ or $\mathbb{R}^+$ by doing some transformations. However, none of them satisfies that for any unbounded set $E$, $\{f_n(x)\}$ does not converge uniformly on $E$.

I know the Egorov's theorem, it states that pointwise convergence in a finite measure space implies "almost" uniform convergence. So this question could be considered as a counterexample that for an infinite measure space, for example $\mathbb{R}$, we may not find a very "large" subset in which converges uniformly.

I wonder if I need to consider rationals to construct such a "strange" function. Thank you for all the help.

Answer 1

The key here use to use functions where the convergence "fails" in $+\infty$.

Consider $$f_{n}(x)= \begin{cases} 0 & x < |n|\\ 1 & x \geq |n|\\ \end{cases}$$

$f_{n}$ converges to 0 pointwise. However, for any unbounded set $E$, and any $n \in \mathbb{N}$. There is $x \in E$ such that $x>n$. So $$\sup\limits_{x\in E}|f_{n}(x)|=1$$

This works as a counter example for Ergorov's theorem when the measure is not finite. Indeed, Ergorov's theorem states

Let $(f_{n})$ be a sequence of $M$-valued measurable functions, where $M$ is a separable metric space, on some measure space $(X,\Sigma,\mu)$, and suppose there is a measurable subset $A \subseteq X$, with finite $\mu$-measure, such that $(f_{n})$ converges $\mu$-almost everywhere on $A$ to a limit function $f$. The following result holds: for every $\varepsilon > 0$, there exists a measurable subset $B$ of $A$ such that $\mu(B) < \varepsilon$, and $(f_{n})$ converges to $f$ uniformly on $A \setminus B$.

If we take $\mathbb{R}$ with the Lebesgue measure, any set $B$ of finite measure must have $\mathbb{R} \setminus B$ unbounded, otherwise $\mathbb{R}$ would be of finite measure. So Ergorov's theorem would fail in this case.