I need some help with this exercise:
If we have $N=2$, $\Omega=\mathbb{R}^2$ and $T:\cal{D}(\Omega)\to\mathbb{C}$, with
$$\langle T,\phi\rangle=\phi(0,1)-\phi(1,0)$$
I have to show that it is a Radon measure, but not a locally integrable function.
Notes: $\bullet$ $T: C_c(\Omega)\to \mathbb{C}$ linear and continuous is a Radon measure, and we denote it by $T_{\mu}$,
$<T_{\mu},\phi>$=$\int \phi\;d\mu$
$\bullet$ $C_c(\Omega)$ is the set of continuous functions from $\Omega$ to $\mathbb{C} $ that have as support a compact of $\Omega$.
Any ideas?
$\delta_{(x,y)}(\varphi)=\varphi(x,y)$ by definition. And probably you have shown already that $\delta_{(0,0)}$ is a Radon measure.
Assume that $f\in L^1_{loc}$ is a locally integrable function inducing $\delta$, for simplicity assume $\Omega=R$ in dimension 1. Let $\varphi$ be a test function with $\varphi(0)=1$, supp$(\varphi)\subset[-1,1]$ and consider $\varphi_\varepsilon(x):=\varphi(x/\varepsilon)$:
$$ 1=\delta(\varphi_\varepsilon)=\int f\varphi_\varepsilon = \int_{-\varepsilon}^{\varepsilon} f(x) \varphi(x/\varepsilon) dx \leq \sup|\varphi| \int_{-\varepsilon}^{\varepsilon} |f| $$
As $f$ is assumed locally integrable, the right hand side converges to 0 as $\varepsilon\to 0$, a contradiction.