Of course the Dirac delta is not a function. Despite, I think the concept of a measure is much easier than that of a distribution. Therefore, I was wondering: In what sense is the concept of a Dirac distribution equivalent to the Dirac measure? Are you (in principle) able to prove all the properties of the distribution if you are using the concept of a measure? Or is the only thing that the Dirac delta measure is good for to say:
$$\int_{\mathbb{R}} f(x)\delta(x-x_0) d\mu(x):= \int_{\mathbb{R}} f(x)d\delta_{x_0}=f(x_0)?$$
Or differently: Would this definition be an appropriate definition? Or do we have to refer to the theory of distributions to prove all the properties that the Dirac-delta has?
Best Answer
The Dirac distribution is not the Dirac measure (point measure), but induced by the Dirac measure. Every Radon measure $\mu$ on $\mathbf{R}$ induces a distribution by $$\phi\mapsto \int_{\mathbf{R}}\phi \ d\mu.$$ Yes we can define the distribution $\delta_0\in\mathscr{D}'(\mathbf{R})$ to be the one induced by the Dirac measure, or simply by $\delta_0(f)=f(0)$. These are obviously the same distribution. I don't quite get your second question, if you consider the Dirac distribution as a distribution (and as such it is properly viewed) you cannot get around distribution theory. Apart from that distribution theory is not that complicated (apart from, say, the topology on $\mathscr{D}$ but according to Hörmander you don't need to understand it) and this is one reason why it is so successful.