Showing upper semicontinuity of $u^*(x) = \lim\limits_{r\to 0}\left( \sup\limits_{B_r(x)}u \right)$. Morning.
I had a similar post yesterday but I noticed after a good sleep tonight that whad I did there does not work. As the post says I want to show that $u^*(x) = \lim\limits_{r\to 0}\left( \sup\limits_{B_r(x)}u \right)$ is upper semicontinuous for arbitrary $u:U\to\mathbb{R}^n$ with $U\subset\mathbb{R}^n$ .
For arbitrary $y\in B_r(x) \subset U$ there is
$u(y) \leq \sup\limits_{B_r(x)}u$
and thus
$u(y) \leq \lim\limits_{r\to 0}\left( \sup\limits_{B_r(x)}u \right)$ for all $y\in B_r(x)$.
Now set $\delta := r-|y-x|$ to obtain
$\lim\limits_{\alpha\to 0}\left( \sup\limits_{B_\alpha(y)}u \right) \leq \sup\limits_{B_\delta(y)}u \leq \lim\limits_{r\to 0}\left( \sup\limits_{B_r(x)}u \right)$ from above.
As $y\in B_r(x)$ was arbitrary we get
$\limsup\limits_{y\to x}\left( \lim\limits_{\alpha\to 0}\left( \sup\limits_{B_\alpha(y)}u \right)\right) \leq \sup\limits_{y\in B_r(x)}\left( \lim\limits_{\alpha\to 0}\left( \sup\limits_{B_\alpha(y)}u\right) \right) \leq \lim\limits_{r\to 0}\left( \sup\limits_{B_r(x)}u \right) = u^*(x)$
and thus $u^*$ is upper semicontinuous.
I hope this is correct now but to be sure: is it?
Thanks along for any help!
Edit: I see that it is not true at all unfortunately as "For arbitrary $y\in B_r(x) \subset U$ there is
$u(y) \leq \sup\limits_{B_r(x)}u$
and thus
$u(y) \leq \lim\limits_{r\to 0}\left( \sup\limits_{B_r(x)}u \right)$ for all $y\in B_r(x)$." doesn't make any sense...but how could I avoid this problem with a neighbourhood of $x$ to work on that doesn't give me problems when $r$ goes to $0$?
 A: It's probably easiest to work with sequences.
You want to show $$\limsup_{x\to x_{\infty}}{u^*(x)}\leq u^*(x_{\infty})$$ for any choice of $x_{\infty}$.  Pick $\{x_n\}_{n=0}^{\infty}\to x_{\infty}$ such that $$\lim_{n\to\infty}{u^*(x_n)}=\limsup_{x\to x_{\infty}}{u^*(x)}$$  Also pick $\{r_n\}_{n=1}^{\infty},\{\epsilon_n\}_{n=1}^{\infty}$ so that $\lim_{n\to\infty}{r_n}=\lim_{n\to\infty}{\epsilon_n}=0$; we'll use these momentarily.
For each $n$, $$u^*(x_n)=\lim_{r\to0}{\sup_{y\in B(x_n,r)}{u(y)}}$$  This limit is decreasing in $r$, so $$u^*(x_n)\leq\sup_{y\in B(x_n,r_n)}{u(y)}$$ (indeed, that inequality would be true for any choice of $r$).  Likewise, there exists $y_n\in B(x_n,r_n)$ such that $$u^*(x_n)\leq u(y_n)+\epsilon_n$$  Varying $n$, $$\limsup_{x\to x_{\infty}}{u^*(x)}\leq\limsup_{n\to\infty}{u(y_n)+\epsilon_n}=\limsup_{n\to\infty}{u(y_n)}$$
On the other side, $$u^*(x_{\infty})=\lim_{r\to0}{\sup_{y\in B(x_{\infty},r)}{u(y)}}$$  Fix $r$ for a moment.
Now, $\lim_{n\to\infty}{y_n}=x_{\infty}$, by the "generalized sandwich theorem": $$|y_n-x_{\infty}|\leq r_n+|x_n-x_{\infty}|\to0$$  Thus, there exists $N$ such that, for $n\geq N$, $y_n\in B(x_{\infty},r)$.  Choosing those points, $$\sup_{y\in B(x_{\infty},r)}{u(y)}\geq\sup_{n\geq N}{u(y_n)}\geq\limsup_{n\to\infty}{u(y_n)}$$  Varying $r$, \begin{align*}
u^*(x_{\infty})&\geq\limsup_{n\to\infty}{u(y_n)} \\
&\geq\limsup_{x\to x_{\infty}}{u^*(x)}
\end{align*} as desired.
