# 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).

• The Dirac delta is not a function in the usual sense. It is what's called a distribution. Jul 3 at 5:13
• This is not a formal definition but rather some intuitive explanation.
– PC1
Jul 3 at 5:26
• – PC1
Jul 3 at 5:27
• The definition is absurd, yes, although it might provide some intuition or some motivation for the rigorous definition. Jul 3 at 5:27
• Please consider writing up the excerpt in your question, not pasting a screenshot of text, as using a screenshot that makes the key part of your question unsearchable and impossible for text-to-voice devices (used by the visually impaired) to understand.
– KCd
Jul 3 at 5:49

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.
• +1. Comments: a usual notation is $\delta_{x_0}$ for the shifted Dirac delta measure you define ;) As I explained in my answer here math.stackexchange.com/questions/3801916/…, a better notation for integrals with another measure is $\int f(x)\,\mu(\mathrm d x)$. With these notations, one could write indeed $$\int f(x) \,\delta_{x_0}(\mathrm d x) = f(x_0)$$ Jul 6 at 10:26