Making a function in $W^{1,2}$ continuous

Question

Let $\Omega$ be an open domain in $\mathbb{R}^n$, $u\in W^{1,2}(\Omega)$ and assume that for any $y$ in $\Omega$

$$\lim_{\varrho \to 0} \operatorname{osc}(u,B(y,\varrho)) \rightarrow 0 , \varrho \rightarrow 0$$ where $$\operatorname{osc}(u,B(y,\varrho))= \sup_{x\in B(y,\varrho)} u(x)-\inf_{x\in B(y,\varrho)} u(x)$$

how can I prove that $${\bar{u}}(x)=\lim_{\varrho \rightarrow 0} \frac{\int_{B(x,\varrho)} u(y) dy }{\operatorname{meas} B(x,\varrho)}$$

is continuous in $\Omega$ ?

EDIT: I am always referring to essential suprema, infima and oscillations, i.e. ignoring sets of measure zero.

Answer 1

OK, I will try myself, please let me know what you think of this...

Essentially we have $$\bar u (x) =\lim_{\varrho\rightarrow 0 } \sup_{t\in B(x,\varrho)} u(t) = \lim_{\varrho\rightarrow 0 } \inf_{t\in B(x,\varrho)} u(t)$$

then let $\epsilon > 0$ and $|x-y|< \delta$

$\sup_{B(y,\delta)} u(t) \geq \bar u(x)$

But also:

$$-\bar u (y) =-\lim_{\varrho \rightarrow 0}\inf_{B(y,\varrho)} u(t)=\lim_{\varrho \rightarrow 0}\sup_{B(y,\varrho)} -u(t)\leq\sup_{B(x,\delta)}-u(t)= -\inf_{B(x,\delta)} u(t) $$

Hence $$\bar u (x) -\bar u (y) \leq \sup_{B(y,\delta)} u(t) -\inf_{B(x,\delta)} u(t)\leq osc(u,B(x,2\delta)) $$

and I can make the RHS as small as I want by assumption, in particular choosing small $\delta$ we can make it small than epsilon to give continuity.