Limit on $L^{\infty}$ space

Question

Suppose that $u_k\to u$ in $L^{\infty}(\mathbb{R}^n)$, and $u_k\geq 0$ for all $k\in \mathbb{N}$ How can I prove that $u\geq 0$ using the $L^{\infty}$ norm? Any help will be appreciated.

Answer 1

If the set such that $u<0$ has zero measure, then $u$ is equal (in $L^{\infty}(\mathbb{R}^{n}),$ since we identify functions that are a.e. equal) to a version which is nonnegative. Otherwise, the set $u<0$ has positive measure, and since $\{u<0\}=\bigcup_{n\geq1}\{u<-1/n\},$ and $\lim_{n\rightarrow\infty}\lambda(\{u<-1/n\})\rightarrow\lambda(\{u<0\})>0,$ there is some positive $n$ such that $\lambda(\{u<-1/n\})>0.$ But on this set, $\sup_{x\in\{u<-1/n\}}|u_{n}(x)-u(x)|>1/n,$ so $\mathrm{ess}\sup_{x\in\mathbb{R}^{n}}|u_{n}(x)-u(x)|>1/n$ and thus $u_{n}\not\rightarrow u$ in $L^{\infty}(\mathbb{R}^{n}),$ which is a contradiction.