# Equivalence between Dirac Delta definition as a measure and as a distribution.

I always thought of Dirac Delta as the distribution $$\delta_ {x_0}$$ which performs $$\phi\ \mapsto \phi(x_0)$$. With respect to this definition we can think the Delta as the weak limit of some $$L^1_{loc}$$ functions (locally integrable) or as the Fourier antitranform of the distribution associated with the constant 1.

At the same time I knew Dirac Measure in $$x_0$$ as the one which assigns to any set of a given sigma-algebra the value of 1 if $$x_0 \in A$$ and 0 otherwise. This is the definition which I think formalize, through Lebesgue integral, the (physicists') integral pseudo-definition of a function always zero and infinity in a point.

Cana anyone enlighten me on how this two definition may be considered equivalent and thus defining some precise mathematical object? Or it is the case that the two are linked but somehow Dirac Delta is a "polymorphic" entity?

Every measure induces a distribution via rule $$\langle \mu, \phi\rangle = \int \phi\ d\mu$$ (converse isn't true - not every distribution is induced by some measure). $$\delta$$ (as distribution) is induced by some measure, and this measure is often itself denoted as $$\delta$$, so this symbol indeed is designated to two formally different objects (function on test functions, and functions on subsets of say $$\mathbb{R}$$).

It's not unique case - there are a lot of objects denoted with symbol $$0$$. And, for example, we define a group as pair $$\langle G, \cdot\rangle$$, but then talk about "elements of group" (thus "element of" sometimes refers to set $$\in$$, and sometimes to something else).

The Dirac Delta is as you mention yourself a distribution and not a function. The integral of $$\int{}f\delta{}d\mu$$ will always equal $$0$$ as you note thus proving that there is no Lebesgue integrable function with the properties of the Dirac Delta. It can only be defined as the linear functional such that $$\int{}f(x)\delta(x-x_0)dx=f(x_0)$$ and does not have a meaning outside the integral.