I just want to confirm the following proof:
Consider a function $u: \Omega \rightarrow \mathbb{R}$ where $\Omega \subset \mathbb{R}^{n}$ and $u \in C^{2}(\bar{\Omega})$. Let $a_{jk}$ be smooth functions on $\bar{\Omega}$. We define function $$\partial^{*}_{\nu}u := \sum_{j,k=1}^{n}a_{jk}(\frac{\partial}{\partial x_{k}}u)\nu_{j}$$ $\nu$ is the outward unit surface normal to $\partial \Omega$, $\nu_{j}$ is the $j$-th component. If we are given that $$\int_{\partial\Omega}u(\partial^{*}_{\nu}u)d\sigma = 0$$ We know that $u(\partial^{*}_{\nu}u) = 0$ almost everywhere, but since $u(\partial^{*}_{\nu}u)$ is continuous I want to prove that $u(\partial^{*}_{\nu}u) = 0$ everywhere.
Proposed Proof:
Since we have $u(\partial^{*}_{\nu}u) = 0$ almost everywhere, there exists some $\sigma$-negligible set $N \subset \partial \Omega$ such that $\sigma(N) = 0$ and where $x \in N \Rightarrow u(\partial^{*}_{\nu}u)(x) > 0$. Then using the continuity of $u(\partial^{*}_{\nu}u)$ we take $\epsilon := |u(\partial^{*}_{\nu}u)|$. By definition of continuity there exists a $\delta > 0$ such that $y \in B(x,\delta) \Rightarrow f(y) \in B(f(x),\epsilon)$. But then since $B(x,\delta)$ is open it follows that $\sigma(B(x,\delta)) > 0$ which is a contradiction. Therefore $u(\partial^{*}_{\nu}u) = 0$ everywhere. $\square$
Is this okay? Please advise on any recommended change. Thanks.
Your assumption that $u(\partial^{*}_{\nu}u) = 0$ almost everywhere does not follow that easily by continuity. You would also require that $u(\partial^{*}_{\nu}u) = 0 \geq 0$. Consider counterexample where we have the continuous function $f(x) = \text{sin}(x)$, take $\Omega = [-\pi, \pi]$ then $\int_{\Omega}f d\lambda = 0$ and certainly $f \neq 0$ a.e.
If you have $u(\partial^{*}_{\nu}u) = 0 \geq 0$ then consider the following proof:
If $f$ is a positive continuous function(therefore measurable) and $\int_{\Omega}f d\lambda = 0$ then $f = 0$ almost everywhere.
Proof:
I will denote the characteristic functions as $\chi$. Consider the sequence of measurable sets $E_{n}:= \{ x \in \Omega: f(x) \geq \frac{1}{n} \}$ then $$0 \leq \lambda(E_{n}) = \int_{\Omega}\chi_{E_{n}}d\lambda \leq n\int_{\Omega}f\chi_{E_{n}}d\lambda \leq n\int_{\Omega}f d\lambda = 0$$ for all $n \in \mathbb{N}$. Therefore $\lambda(E_{n}) = 0$ for all $n \in \mathbb{N}$. Since $E_{n} \subset E_{n+1}$ we have $\lambda(\cup_{n=1}^{\infty}E_{n}) = \lim\limits_{n \rightarrow \infty}\lambda(E_{n}) = 0$. Then it is clear that for any $x \in \Omega$ such that $f(x) > 0$ there exists a $m \in \mathbb{N}$ such that $x \in E_{m} \subset \cup_{n=1}^{\infty}E_{n}$. Therefore $f = 0$ almost everywhere. $\square$
The proof you gave seems fine, given that you have assumed that $u(\partial^{*}_{\nu}u) \geq 0$. I think you could even easily note that if $f$ is continuous and $f = 0$ a.e. then $f^{-1}(\mathbb{R}\setminus \{0\})$ is open and since an open set other than the emptyset has Lebesgue measure zero it follows that $f^{-1}(\mathbb{R}\setminus \{0\}) = \emptyset$, so $f = 0$ even everywhere. But I'm assuming you specifically want your proof checked.