# Locally Sobolev space and Sobolev spaces

In Reed, Simon, Methods of Mathematical Physics, it defines the Sobolev space $$W_a, a\in \mathbb{R}$$ as the set of tempered distributions $$\mu\in S'(\mathbb{R}^n)$$ such that its Fourier transform $$\hat{\mu}$$ is measurable and $$|| \mu ||_a^2 = \int (1+p^2)^a |\hat{\mu}|^2 dp <\infty$$ Let $$\Omega \subseteq \mathbb{R}^n$$ be open. Then the local Sobolev spaces on $$W_a (\Omega)$$ is the set of distributions $$\mu \in D'$$ such that $$\phi \mu \in W_a$$ for every $$\phi \in D(\Omega) = C_c^\infty(\Omega)$$.

It seems that if $$\Omega = \mathbb{R}^n$$, then the local Sobolev space $$W_a (\Omega)$$ should equal the Sobolev space $$W_a$$. However, I can't seem to prove it. Any help would be appreciated.

This is false. In the same way, locally integrable functions are not necessarily integrable on $$\mathbb{R}^d$$. You need some decay at infinity to be in global Sobolev spaces.
Take for example $$μ = 1$$. Then for any $$\varphi\in C^\infty_c$$, $$\varphi\,\mu = \varphi\in W_a$$. But of course $$\mu\notin L^2$$ so $$\mu\notin W_a$$.