# Radon measure not locally integrable

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:

• $$T: C_c(\Omega)\to \mathbb{C}$$ linear and continuous is a Radon measure, and we denote it by $$T_{\mu}$$,

$$\langle T_{\mu},\phi\rangle=\int \phi\;d\mu$$

• $$C_c(\Omega)$$ is the set of continuous functions from $$\Omega$$ to $$\mathbb{C}$$ that have as support a compact of $$\Omega$$.

Any ideas?

• So you are talking about the distribution $\delta_{(0,1)} - \delta_{(1,0)}$. What do you know about this one?
– Vobo
Commented Nov 8, 2013 at 7:24
• I'm a starter at distribution theory. I haven't met this kind of distribution or exercise before...I'm editing the post with my definition of Radon measure. Commented Nov 8, 2013 at 7:36
• And Vobo, $\delta$ function is not involved here, I think. $\phi$ denotes any funtion in $D(\Omega)$ (functions of class $C^{\infty}$ in $\Omega$, and with compact support) Commented Nov 8, 2013 at 7:46

$\delta_{(x,y)}(\varphi)=\varphi(x,y)$ 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.

• Thanks, Vobo. In the case $N=2$ where we have the difference of $\phi$'s, as in the statement, how could we adapt your reasoning? Maybe choosing $\varphi$ test function with $\varphi(0,1)=1$ and $\varphi(1,0)=0$ , with $supp(\varphi)\subset [0,1.5]$X$[0,1.5]$ or something like this? Commented Nov 8, 2013 at 13:34
• The support of the choosen $\varphi$ should contain only one of these two points.
– Vobo
Commented Nov 8, 2013 at 15:40
• Ok, I think I have it. If I define $\varphi$ with support contained in,for example $[-0.5,0.5]$x$[-2.2]$, and verifying that $\varphi(0,1)=1$, then if we consider $\varphi_{\epsilon}(x)=\varphi(x/\epsilon,y)$, we can use an argument similar to yours, leading us to $1-0=\delta_{(0,1)}(\varphi_\epsilon)-\delta_{(1,0)}(\varphi_\epsilon)=\varphi(0,1)-\varphi{(1/\epsilon,0)}=\int_{-0.5\epsilon}^{0.5\epsilon}\int_{-2}^{2}f(x,y)\varphi(x/\epsilon,y) dy dx\leq sup|\varphi|\int_{-0.5\epsilon}^{0.5\epsilon}\int_{-2}^{2}|f|$. Taking limits, we have the contradiction. Is this ok? Commented Nov 8, 2013 at 19:16
• Yes, looks good to me.
– Vobo
Commented Nov 8, 2013 at 21:21