Show that $M\subseteq\bigcup\limits_{m=1}^{\infty}\bigcap\limits_{n=1}^{\infty}\bigcup\limits_{j=n}^{\infty}\left\{ f_j(x)-f(x)>\frac{1}{m}\right\}$

Question

Let be $(f_n)_{n\in\mathbb{N}}$ a sequence of functions, with $f_n:\mathbb{R}\to\mathbb{R}$ for all $n\in\mathbb{N}$ and $f:\mathbb{R}\to\mathbb{R}$. We define $M:=\left\{x\in\mathbb{R}\mid \liminf\limits_{n\to\infty}f_n(x)> f(x)\right\}$. Show that \begin{align*} &M\subseteq\bigcup\limits_{m=1}^{\infty}\bigcap\limits_{n=1}^{\infty}\bigcup\limits_{j=n}^{\infty}\left\{x\in\mathbb{R}\mid f_j(x)-f(x)>\frac{1}{m}\right\}. \end{align*}

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Let's choose an arbitrary $x\in M$ and assume $\liminf\limits_{n\to\infty}f_n(x)-f(x)=\delta>0$. WLOG we assume $\delta$ to be rational otherwise we find a $r\in\mathbb{Q}$ such that $0<r<\delta$. We know that $\liminf\limits_{n\to\infty}f_n(x)$ is a limit point so that we find infinitely many $k\in\mathbb{N}$ with $\liminf\limits_{n\to\infty}f_n(x)-f_k(x)<\frac{\delta}{2}$. Hence, we find infinitely many $k\in\mathbb{N}$ such that $f_k(x)-f(x)>\frac{\delta}{2}$. So $$ x\in \bigcup\limits_{m=1}^{\infty}\bigcap\limits_{n=1}^{\infty}\bigcup\limits_{j=n}^{\infty}\left\{x\in\mathbb{R}\mid f_j(x)-f(x)>\frac{1}{m}\right\}. $$

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Is this correct?

Answer 1

Your idea is correct but I believe that in the end you mixed up some indices. So let $x \in M$ and $\liminf_{n \to \infty} f_n(x) = \delta + f(x)$ for some $\delta>0$. This means that a subsequence $(f_{n_j}(x))_j$ converges to $\delta + f(x)$. This means: For every $n \in \mathbb N$ we find a $j \ge n$ with $\lvert f_j(x) - f(x) \rvert > \frac{\delta}{2}$. This shows $$x \in \bigcap_{n=1}^\infty \bigcup_{j=n}^\infty\{x \in \mathbb R: \lvert f_j(x) - f(x) \rvert >\frac{\delta}{2}\}$$

Now there is $m \in \mathbb N$ such that $\frac 1m \le \frac{\delta}{2}$ because $\lim_{m \to \infty} \frac 1m =0$. This implies $$x \in \bigcup_{m=1}^\infty\bigcap_{n=1}^\infty \bigcup_{j=n}^\infty\{x \in \mathbb R: \lvert f_j(x) - f(x) \rvert >\frac{1}{m}\}$$

Answer 2

In fact much more is true: $$M=\bigcup_{m=1}^{\infty}\bigcup_{n=1}^{\infty}\bigcap\limits_{j=n}^{\infty}\left\{x\in\mathbb{R}\,\left|\,f_j(x)-f(x)>\frac1m\right.\right\}.$$

• Proof of $\subset$: let $x\in M.$ Then, for some $m\ge1,$ $\liminf f_n(x)-f(x)>\frac1m,$ i.e. $$\exists n\ge1\quad\forall k\ge n\quad\inf_{j\ge k}\left(f_j(x)-f(x)\right)>\frac1m.$$ This implies $\forall j\ge n\quad f_j(x)-f(x)>\frac1m.$
• Proof of $\supset$: let $x\in\Bbb R$ be such that for some $m,n\ge1,$ for all $j\ge n,$ $f_j(x)-f(x)>\frac1m.$ Then $$\forall k\ge n\quad\inf_{j\ge k}\left(f_j(x)-f(x)\right)\ge\frac1m.$$ This implies $\liminf f_n(x)-f(x)\ge\frac1m,$ hence $x\in M.$
• Proof that the equality above is "much more" than the requested inclusion: for any sequence $(A_n)_n$ of sets, $$\bigcup_{n=1}^{\infty}\bigcap\limits_{j=n}^{\infty}A_j=\liminf A_n\subset\limsup A_n=\bigcap_{n=1}^{\infty}\bigcup\limits_{j=n}^{\infty}A_j.$$