Assume we are in $\mathbb R^N$ and $\Gamma$ is a ($N-1$)-rectifiable set with $\mathcal H^{N-1}(\Gamma)<\infty$ and $\mathcal H^{N-1}(\bar \Gamma\setminus \Gamma)=0$. Let $u\in C_c(\mathbb R^N)$ and $u\geq 0$.
Define $\Gamma_\epsilon = \{x\in\mathbb R^N\mid \operatorname{dist}(x,\Gamma)<\epsilon\}$. I am trying to prove that $$ \lim_{\epsilon\to 0}\frac{1}{2\epsilon}\int_{\Gamma_\epsilon}u(x)\,dx = \int_\Gamma u(x)\,d\mathcal H^{N-1}. $$
I feel like this is similar to Lebesgue point theorem, but instead of comprising to a point, it goes to a $N-1$ rectifiable set.
No.
Let $\Gamma$ be the collection of rational points inside the unit ball and suppose that $u = 1$ on a neighborhood of the unit ball. Since $\Gamma$ is countable it is trivially $N-1$ rectifiable, and since $\Gamma_\epsilon$ contains the unit ball for $\epsilon > 0$ you have $\displaystyle \int_{\Gamma_\epsilon} u(x) \, dx \ge 1$. The limit on the left is infinite. On the other hand, $\Gamma$ has Hausdorff dimension $0$ so that $\mathcal H^{N-1}(\Gamma) = 0$ and the integral on the right equals $0$.
You may wish to modify the question and require $\Gamma$ to be compact.