While studying PDE, I came across this trace operator which talks about the class $L^{p}(\partial \Omega)$ for open sets $\Omega \subset \mathbb{R}^{n}$ with $C^{1}$ boundary.
I don't understand what measure we give to $\partial \Omega$.
[Math] Meaning of surface measure
definitionmeasure-theorypartial differential equationstrace-map
Related Solutions
What Anthony Carapetis wrote in his answer, the weak solution, is still in distributional sense, namely, the Poisson equation $-\Delta u = f$ holds in $H^{-1}$.
For the smooth case, we don't have to go into the realm of weak solution.
The uniqueness of the solution, given that a solution exists, is obtained by maximum principle (see Evans 2.2.3):
If $\Omega$ is open and bounded (equivalent to your assumption that $\overline{\Omega}$ is comapct), if $\Delta u = 0$ in $\Omega$, for some $u$ is at least twice continuously differentiable in $\Omega$, and continuous up to the boundary of $\Omega$. Then $$ \max_{x\in \Omega} u(x) = \max_{x\in \partial\Omega} u(x). $$
Replacing $u$ with $-u$ we will see that $$ \min_{x\in \partial\Omega} u(x)\leq u(x) \leq \max_{x\in \partial\Omega} u(x). $$ Thus $\Delta w = 0$ and $w|_{\partial\Omega}=0$ give us $w=0$. Where this $w$ here is the difference between two smooth functions satisfying the same boundary value problem.
For existence, the construction is actually explicit! This is called Green's representation formula, please refer to Evans 2.2.4 and Theorem 12.
But the thing is the construction relies on the existence of $\phi(x,y)$ such that for a fixed $x\in \Omega$ $$ \begin{cases}\Delta_y \phi(x,y) = 0 & \text{in }\Omega, \\[4pt] \phi(x,y) = \Phi(y-x) &\text{on }\partial \Omega, \end{cases} \tag{$\star$} $$ where the $\Phi(y-x)$ is the Green function solving the $-\Delta u = \delta_x$ on the whole $\mathbb{R}^n$. If we can prove the existence of a solution for above boundary value problem of Laplace equation $(\star)$, we are done.
Moreover $u$ can be represented as: $$ u(x) = \int_{\Omega} f(y)G(x,y)dy - \int_{\partial \Omega} g(y) \frac{\partial G}{\partial n}(x,y)dS(y), $$ where $\partial G/\partial n = \nabla G\cdot n$ is the directional derivative, and $G(x,y) = \Phi(x-y)-\phi(x,y)$. This $u$ solves $$ \begin{cases}-\Delta u = f & \text{in }\Omega, \\[4pt] u = g &\text{on }\partial \Omega, \end{cases} $$ for $f$ and $g$ satisfying certain smoothness.
The existence of a solution for $(\star)$ is a difficult analysis problem, when $\partial \Omega$ is relatively "simple", we can transform the domain into a ball which has an explicit expression Green function. For the existence in general bounded, smooth domain, we can use Perron's method of subharmonic functions (See Gilbarg-Trudinger 2.8, exercise 2.10).
Short answer: The measure is the higher dimensional analogue of arclength and surface area. It's not Lebesgue measure from the ambient Euclidean space directly. But $\mathbb{R}^N$ is an inner product space, and that is used to determine lengths of vectors, volume of boxes, and Lebesgue measure on $\mathbb{R}^N$. Any submanifold of $\mathbb{R}^N$ inherits that inner product in the form of a Riemannian metric, which determines lengths of tangent vectors, volume of “infinitesimal” (i.e. tangent) boxes, and by integrating, a measure.
At more length:
No, the measure on the boundary $\partial\Omega$ is not Lebesgue measure on $\mathbb{R}^{N-1}$. If $\Omega$ is an $N$-dimensional domain in $\mathbb{R}^N$, then its boundary $\partial\Omega$ is an $(N-1)$-dimensional submanifold of $\mathbb{R}^N.$ It is not a subset of $\mathbb{R}^{N-1}$, so its measure cannot be computed using Lebesgue measure on $\mathbb{R}^{N-1}$.
In general, a $k$-dimensional submanifold of $\mathbb{R}^N$ need not be naturally a subset of any $\mathbb{R}^k$, so we can't use Lebesgue measure on $\mathbb{R}^k$ to measure it.
However, a $k$-manifold is locally homeomorphic to $\mathbb{R}^k$, which means that each local patch has a $\sigma$-algebra isomorphic to that of $\mathbb{R}^k$. Does this give us a measure we can use for the $k$-dim submanifolds of $\mathbb{R}^N$? No, a submanifold may be topologically nontrivial, like a 2-sphere, so that it requires more than one coordinate patch to cover it. A set may be in more than coordinate patch, and what measure you assign it depends on which coordinates you use. So that's not well-defined.
But to state the problem is to solve it. When you change coordinate patches, the measure as defined in $k$-dimensional Euclidean space transformed by the Jacobian of the transition map. If you can account for that transformation, you will get a well-defined measure.
Here it is in more detail, but what follows is a straightforward generalization of the notions of arc-length and surface area from multivariable calculus, using a bit of the language of linear algebra and Riemannian geometry.
First, the linear algebra. Given an inner product space $V$, we assign a vector $v$ a length $\lVert v\rVert = \sqrt{\langle v,v\rangle}.$ We represent parallelograms spanned by two vectors $u,v$ by their wedge product $u\wedge v$ in the exterior algebra on $V$. We can extend the inner product to these paralellograms (also known as bivectors or 2-planes):
$$\langle u\wedge v, x\wedge y\rangle=\langle u,x\rangle\langle v,y\rangle-\langle u,y\rangle\langle v,x\rangle.$$
Then the area of a parallelogram is the magnitude of the corresponding bivector:
$$\text{area}(u,v)=\lVert u\wedge v\rVert = \sqrt{\langle u\wedge v,u\wedge v\rangle}.$$
Similarly, for any $k<N$, we consider $k$-dimensional paralellepipeds aka $k$-planes, which we represent as $k$-vectors, which are wedges of $k$ different vectors. We have the general formula for the inner product of $k$-vectors as a Gram determinant:
$$\langle u_1\wedge\dotsc\wedge u_k, v_1\wedge\dotsc\wedge v_k\rangle=\det[\langle u_i,v_j\rangle]$$
and
$$\text{vol}_k(u_1,\dotsc,u_k)=\lVert u\wedge \dotsb\wedge u_k\rVert = \sqrt{\langle u\wedge \dotsb\wedge u_k,u\wedge \dotsb\wedge u_k\rangle}.$$
Now that we know how to compute the volumes of boxes, we slice our $k$-dimensional manifold into infinitesimal pieces, compute the size of each piece as the volume of the $k$-plane spanned by the tangent vectors. We sum the volumes, Riemann integral-style, to get the total measure. In other words, if our $k$-manifold $M$ is parametrized as $\mathbf{x}(u_1,u_2,\dotsc,u_k)$, the tangent space is spanned by $\partial\mathbf{x}/\partial u_1,\dotsc, \partial\mathbf{x}/\partial u_k$. The infinitesimal tangent box has volume $$\lVert\partial\mathbf{x}/\partial u_1\wedge\dotsb\wedge\partial\mathbf{x}/\partial u_k\rVert$$ and the measure of the submanifold is given by
$$\text{vol}_k(M) = \int_M\lVert\partial\mathbf{x}/\partial u_1\wedge\dotsb\wedge\partial\mathbf{x}/\partial u_k\rVert\,du_1\,\dotsc\,du_k.$$
And that's the measure. Put $k=N-1$ to compute the measure of the $(N-1)$-dimensional submanifold $\partial\Omega\subseteq\mathbb{R}^N$.
To clarify some terminology, any function which assigns numbers to $k$-planes (and subject to certain transformation rules to ensure it's well-defined which make it a density ) can be integrated, Riemann integral style. If these functions are also linear, they are called differential forms, but the ones we're considering are not linear, so they're only densities.
The integral of the density gives the measure.
Best Answer
Notice $\partial \Omega$ is a Riemannian hypersurface of $\mathbb R^n$. The measure on $\partial \Omega$ is the one given by the Riemannian metric.
Let me define the measure more explicitly.
Since $\Omega$ is open (and therefore orientable), $\partial \Omega$ is orientable. Knowing $\partial \Omega$ is an orientable hypersurface we have a global normal vector field $n$. We can suppose $|n(p)|=1$ for all $p\in \partial \Omega$.
For each $p$ define $$\eta(v_1,\ldots,v_{n-1}) = \det(n(p),v_1,\ldots,v_{n-1})$$ where $v_1,\ldots,v_{n-1} \in T_p \partial \Omega$. This $\eta$ is the Riemannian volume form of $\partial \Omega$.
The measure you want is the one provided by this volume form. More precisely you can obtain the measure using the Riesz–Markov–Kakutani representation theorem on the functional $C(\partial \Omega) \to \mathbb R$ given by $$f \mapsto \int_{\partial \Omega}f \eta.$$