Let $E$ be a normed $\mathbb R$-vector space, $B_r:=\{x\in E:\left\|x\right\|_E<r\}$ for $r>0$, $\mu$ be a measure on $\mathcal B(E)$ with $\mu(E)=\infty$, $p\in[1,\infty)$ and $f\in\mathcal L^p(\mu)$.
Are we able to show that $$\forall\varepsilon>0:\exists r>0:\sup_{x\in E\setminus B_r}\left|f(x)\right|<\varepsilon\tag1?$$
Maybe we can show that, more generally, if $(E,\mathcal E,\mu)$ is any measure space and $\mathcal B\subseteq\mathcal E$ is any system containing a sequence $(B_n)_{n\in\mathbb N}\subseteq\mathcal B$ with $B_1\subseteq B_2\subseteq\cdots$ and $\bigcup_{n\in\mathbb N}B_n=E$, then $$\forall\varepsilon>0:\exists B\in\mathcal B:\sup_{x\in E\setminus B}\left|f(x)\right|<\varepsilon\tag2.$$
The idea is clearly to use that $\mu(B_n)\xrightarrow{n\to\infty}\mu(E)=\infty$. Moreover, we would somehow need to bound $\left\|f\right\|_{L^p(\mu)}^p$ from below by something which tends to $\infty$.
No, that is false. Take $\mu$ to be Lebesgue measure on $\mathbb R$, $p=1$ and $f=\sum_n n \chi_{(n, n+\frac 1 {n^{3}})}$.