How is the Dirac function different from the indicator function

Question

My question is straightforward though I cant find an answer online. What is the difference between an indicator function and a Dirac function?

Answer 1

Let $A=\{0\}$ and indicator function on $A$ be $I_A$. Measure theoretically, $I_A=0$ a.e (almost everywhere) and
$\int_{\mathbb{R}}I_A=0.$
Dirac function is point mass at origin (in fact, it is a functional on set of compactly supported smooth functions) and can not be defined point-wise. Most importantly $\int_{\mathbb{R}}\delta(x)dx=1$ .

Answer 2

The Dirac delta distribution is not an indicator because it is infinite at the point $0$, rather than just taking the value $1$ there. (The indicator function is a bona fide function; the Dirac delta distribution is a distribution, or generalised function.)

If $\delta$ were the indicator function of some set, what set would it be indicating?

Answer 3

In probability theory, given a probability space $\textstyle (\Omega, \mathcal F, \mathbb P)$ with $A \in \mathcal F$, the indicator random variable $\mathbf{1}_A \colon \Omega \rightarrow \Bbb{R}$ is defined by $\mathbf{1}_A (\omega) = 1 $ if $ \omega \in A,$ otherwise $\mathbf{1}_A (\omega) = 0.$ The expected value of $\mathbf{1}_A (\omega)$ is identical to the probability of $A$: $ \mathbb{E}( \mathbf{1}_A (\omega)\,) \equiv \Pr(A) $.

With the Dirac $\delta$ defined in the usual way, and the standard abuse of notation, we can write $$\mathbf1_{A}(x)=\int_A \delta(x-x') \mathrm{d}x'. $$ and get an expression of the indicator function in terms of the delta distribution.