The Dirac measure is defined by $$\delta_x(A)= \begin{cases} 1 &\text{if $x \in A$}\\ 0 &\text{if $x \notin A$}\\ \end{cases}$$ Let $f:X\rightarrow \mathbb{R}$ be a function.
Then we will have: $\int \limits_{X} f \, d\delta_x=f(x)$ $(*)$?

  1. I'm confused about the notation $x$ in $f(x)$ in the $(*)$ equation: which value of $x$ that we should choose while we are computing the integral over the set $X$ which consists of many value of $x$.

  2. What is the role of the set $A$ (defined in the Dirac measure) in the computation of the integration $\int \limits_{X} f \, d\delta_x$

Thank you very much for your explanation!

  • $\begingroup$ $x$ is fixed. This should help you. $\endgroup$ Feb 27, 2021 at 17:03
  • $\begingroup$ @Matematleta: I have read this thread before posing this question indeed. Oh, so in this case, $x$ is fixed and the set $X$ is actually $A$ right ? $\endgroup$ Feb 27, 2021 at 17:05
  • $\begingroup$ No. $x$ is fixed but $A$ s not. $\delta$ is a measure defined on some $\sigma$-algebra of subsets of $X$ of whch $A$ is any member. If $x\in A$ then $\delta$ returns the value $1$; if not then it returns $0.$ Using the definition of the integral (as an increasing limit of simple functions) then gives the result you want to prove. $\endgroup$ Feb 27, 2021 at 17:12

1 Answer 1


In general, for a measure $\mu$, one can find the notations $$ \langle \mu,f\rangle = ∫ f\,\mu = ∫ f(x)\,\mu(\mathrm d x) = \int f\,\mathrm d\mu = \int f(x)\,\mathrm d\mu(x), $$ to denote the integral of a function $f$ with respect to the measure $\mu$. Remark that the $x$ variable is a dummy variable, one can choose whatever name, such as $$ ∫ f\,\mu = \int f(y)\,\mu(\mathrm d y) = \int f(x)\,\mu(\mathrm d x). $$

  1. In your formula, you first fix a $x\in X$ that has nothing to do with the integral. Then you define the measure $μ = \delta_x$. Then you look at the integral with respect to this measure (written in your case with the 3rd convention of my first equation) $$ \int_X f \,\mathrm d\delta_x = ∫_X f(y)\,\delta_x(\mathrm d y) = f(x). $$

  2. A (real-valued) measure can be defined in two equivalent ways thanks to Riesz representation theorem:

  • A functional assigning to each measurable set a value: for every set $A\subset X$, $μ(A)\in\mathbb R$.
  • A functional assigning to each continuous function a value: for every function $f$, $\mu(f) = ∫_X f\,\mu$.

A way to understand how to go from one to the other point of view is to remark that if you have a sequence of continuous functions $f_n$ converging to the characteristic function of a set $A⊂X$, then $$ \int_X f_n(x)\,\mu(\mathrm d x) \to \int_X \mathbf 1_{A}(x) \, \mu(\mathrm d x) = \int_A \mu(\mathrm d x) = \mu(A), $$ while, in the other direction, if you are given a measure $\mu$ as a functional on every set $A⊂ X$, you can build the integral with respect to this measure in a similar way a one build the classical integral with respect to the Lebesgue measure. So, one assigns the value $$ \int_X \mathbf 1_{A}(x) \, \mu(\mathrm d x) := \mu(A), $$ and then one can approximate measurable functions by sum of characteristic functions of sets.

In the case of the Dirac delta, one has the equivalence between your definition $\delta_a(A) = 1$ if $a∈A$ and $0$ if not, and the integral definition $$ ∫ f \,\delta_a = f(a). $$

Remark: I do not recommend the notations $\int f\,\mathrm d \mu = \int f(x)\,\mathrm d \mu(x)$ which are incoherent with the Stieltjes integral notation, in which one writes $μ = \mathrm d g$ for a function $g$ that is the derivative of the measure of $\mu$, so $$ ∫ f(x)\,\mu(\mathrm d x) = ∫ f(x)\,\mathrm dg(x). $$


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