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Let $d\in\mathbb N$, $\Lambda\subseteq\mathbb R^d$ be bounded and open and $p\ge 1$. In the definition of the trace operator, we consider the space $L^p(\partial\Lambda)$? But since $\partial\Lambda$ has Lebesgue measure zero and $L^p(\partial\Lambda)$ is a set of equivalence classes of Lebesgue almost equal functions, shouldn't it contain only the equivalence class of the zero function $\partial\Lambda\to\mathbb R$?

0xbadf00d
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  • When you say $\partial \Lambda$ has Lebesgue measure zero, you mean $\lambda_{\mathbb{R}^d}(\partial \Lambda) = 0$. Which is correct, but here $L^p(\partial\Lambda)$ is equipped with the Lebesgue measure on $\partial \Lambda$ and of course $\lambda_{\partial \Lambda}(\partial \Lambda) = 1$. – anonymus Aug 31 '16 at 18:17
  • @anonymus If $\mathcal B(\mathbb R^d)$ denotes the Borel $\sigma$-algebra and $λ$ the Lebesgue measure on $\mathbb R^d$, then the Lebesgue measure on $∂\Lambda$ is the trace $\left.λ\right|{∂\Lambda}$ of $λ$ on $∂\Lambda$, $$\left.λ\right|{∂\Lambda}:=\left.λ\right|{\left.\mathcal B(\mathbb R^d)\right|{∂\Lambda}}$$ where $$\left.\mathcal B(\mathbb R^d)\right|{∂\Lambda}:=\left{A\cap ∂\Lambda:A\in\mathcal B(\mathbb R^d)\right}=\mathcal B(∂\Lambda)$$ (note that $∂\Lambda\in\mathcal B(\mathbb R^d)$). So, $$\left.λ\right|{∂\Lambda}(∂\Lambda)=λ(∂\Lambda)=0;.$$ What am I missing? – 0xbadf00d Aug 31 '16 at 18:41

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