In the context of vicosity solutions for fully nonlinear PDE of second order (I don't know first order theory) it is useful to define the upper semicontinuous envelope of $u : \Omega \longrightarrow \mathbb R$ as $$ u^* (x) = \lim_{r \to 0} \sup_{B_r(x)}u. $$
To prove that $u^*$ is upper semicontinuous I proceeded as follows:
Fix $x_0 \in \Omega$. $$ \limsup_{x \to x_0} u^*(x) = \lim_{\delta \to 0} \sup_{B_\delta(x_0)\setminus \{x_0\}}\left( \lim_{r \to 0} \sup_{B_r(x)}u \right) \\ \leq \lim_{\delta \to 0} \left(\sup_{B_\delta(x_0) \setminus \{x_0\}} \sup_{B_\delta(x_0)} u\right) \\ = \lim_{\delta \to 0} \sup_{B_\delta(x_0)} u \\ = u^*(x_0), $$ the main point of the argument being that if $r$ is small enough then $B_r(x) \subset B_\delta(x_0)$ and therefore $\sup_{B_r(x)} \leq \sup_{B_r(x_0)} u$.
To prove that $u^*$ is the smallest, take some $v \in USC$, $v \geq u$ and fix $x_0$. We want to prove that $u^*(x_0) \leq v(x_0)$. But $$ u^*(x_0) = \lim_{r \to 0} \sup_{B_r(x_0)} u \leq \lim_{r \to 0} \sup_{B_r(x_0)} v \\ \leq \max\{\liminf_{x \to x_0}v, v(x_0)\} \\ = v(x_0), $$ by the upper semicontinuity of $v$.
Is the above correct?
Thanks in advance.