Does defining Dirac delta via $\int_{-\infty}^\infty f(x)\delta(x-x_0)\ dx = f(x_0)$ make sense? I know Dirac delta can be defined rigorously with distributions, but is the following definition absurd?



I think so since the integral should always be 0 in the sense of Lebesgue (it takes only one positive value, but on a singleton).
 A: In the way it is written, it is absurd. But it can be understood in the following way. If you define the Dirac measure $\delta$ as
$$
\delta(E)=\begin{cases}1,&\ x_0\in E\\[0.3cm] 0,&\ x_0\not\in E\end{cases}
$$
then it is easy to check that the Lebesgue integral for any $f$ is
$$\tag1
\int_{\mathbb R}f(x)\,\delta(x)=f(x_0).
$$
The notation that physicists use comes from the Radon-Nikodym Theorem: if $\mu$ is absolutely continuous with respect to Lebesgue measure, then there exists a function $g$ such that
$$\tag2
\int_{\mathbb R}f(x)\,d\mu(x)=\int_{\mathbb R}f(x)\,g(x)\,dx.
$$
Of course the Dirac measure is not absolutely continuous with respect to Lebesgue measure. But physicists only use the Dirac delta in the equality $(1)$, never as a standalone function, so they don't run into contradictions.
A: Within the theory of integration the definition doesn't make sense. You can even call it absurd.
But it common, especially among physicists, to abuse or extend the integral notation (call it whatever you want) to cover the application of a distribution on a test function (which is a kind of generalization of an integral), and as such it makes sense.
