Sets and pointwise convergence

Question

Let $\{f_n\}_{n\in\mathbb{N}}$ be a sequence of measurable functions which converge pointwise to $f$ in $X\ne\emptyset$. Let $\epsilon>0$ be some constant. We define $$A_m^{\epsilon}=\{x : \text{sup}_{n\ge m}|f_n(x)-f(x)|>\epsilon\}.$$ I have proved that $A_{m}^{\epsilon}\supset A_{m+1}^{\epsilon}$ for all $m\in\mathbb{N}$. How can I show that $$\lim_{m\to\infty} A_m^{\epsilon}=\emptyset.$$ Thanks!

Answer 1

By showing that $\limsup A_m^{\epsilon}=\varnothing$.

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Then based on $\liminf A_m^{\epsilon}\subseteq\limsup A_m^{\epsilon}$ we are allowed to conclude that $$\liminf A_m^{\epsilon}=\varnothing=\limsup A_m^{\epsilon}$$

That allows us to conclude that $\lim A_m^{\epsilon}$ is well defined and takes value $\varnothing$.

It comes to proving that for every $x$ then is an integer $m_x$ such that $m\geq m_x$ implies that $x\notin A_m^{\epsilon}$.

That means that the set $\{m\mid x\in A_m^{\epsilon}\}$ is finite hence $x\notin\limsup A_m^{\epsilon}$.

Answer 2

Every $x\in X$ belongs to only finitely many $A^c_m.$ If not, then for some $x\in X$ we have $\forall m \;\exists n\geq m\;(|f_n(x)-f(x)|>\epsilon /2),$ contradictory to $\lim_{n\to \infty}f_n(x)=f(x).$