Let $(f_{n})_{n\in\mathbb{N}}\subset C(\mathbb{R}^{N})$ be a sequence of real valued function such that $\|f_{n}\|_{L^{\infty}(\mathbb{R}^{N})}\to\infty$ as $n\to\infty$.
Then, I know that there exists $\delta_{1}>0$ and $x_{1}\in\mathbb{R}^{N}$ so that for $n\in\mathbb{N}$ large enough, we have $|f_{n}(x_{1})|\geq\delta_{1}$. Now, I want to show that if I define $X_{\delta_{1}}:=\{x\in\mathbb{R}^{N}\,|\, |f_{n}(x)|\geq\delta_{1} \text{ for }n\text{ large enough}\}$, I want to show that the measure of $X_{\delta_{1}}>0$. Is this actually possible or do I need to add stronger assumption?
I try to derive contradiction by using the fact that $(f_{n})_{n\in\mathbb{N}}$ is a sequence of continuous function but it seems to not work very well. Any help or insight is pretty much appreciated! Thank you!
Let $N=1$ and $$f_n(x)=\begin{cases}n(x-n+1),&\ x\in[n-1,n)\\ n,&\ x\in[n,n+1]\\ n(n+2-x),&\ x\in(n+1,n+2]\\ 0,&\ \text{otherwise}\end{cases}$$ Then $\|f_n\|=n$. But $f_n(x)\to0$ for all $x$. So, for any $k>0$, $$ \{x:\ |f_n(x)|>k\ \text{ for $n$ large enough}\}=\varnothing $$ if you meant "for all $n$ large enough".