# Value of integration with respect to Dirac measure

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!

• $x$ is fixed. This should help you. Feb 27, 2021 at 17:03
• @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 ? Feb 27, 2021 at 17:05
• 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. Feb 27, 2021 at 17:12

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 $$∫ f(x)\,\mu(\mathrm d x) = ∫ f(x)\,\mathrm dg(x).$$