Is the sifting property of the Dirac delta over a finite interval a proven property or a natural interpretation?

Question

I came across the following statement on Wolfram:

The delta function has the fundamental property that $$\int_{-\infty}^{\infty} f(x) \delta(x-a)dx=f(a)$$
and, in fact, $$\int_{a-\epsilon}^{a+\epsilon} f(x) \delta(x-a)dx=f(a)$$ for $\epsilon > 0$.

The first property is well-known and follows from the definition of the Dirac delta as a distribution, where it acts on test functions via $δ(ϕ)=ϕ(0)$. However, I am uncertain about the second property, which restricts the integration to a finite interval $[a−ϵ,a+ϵ]$:

1. Is this a proven property of the Dirac delta, derived rigorously from its definition as a distribution?
2. Alternatively, could this be taken as a definition, justified by a natural interpretation?

Some may argue that this follows intuitively by splitting the integral as:

$$ f(a) = \int_{-\infty}^{\infty} f(x) \delta(x-a)dx = \int_{-\infty}^{a-\epsilon} f(x) \delta(x-a)dx + \int_{a-\epsilon}^{a+\epsilon} f(x) \delta(x-a)dx + \int_{a+\epsilon}^{\infty} f(x) \delta(x-a)dx $$ Then, using the heuristic definition:

$$\delta(x - a)={\begin{cases} \infty &x = a\\ 0,&{\text{otherwise}}\end{cases}}$$

one might claim that $δ(x−a)=0$ outside $[a−ϵ,a+ϵ]$, leading to:

$$ f(a) = \int_{-\infty}^{\infty} f(x) \delta(x-a)dx = \int_{a-\epsilon}^{a+\epsilon} f(x) \delta(x-a)dx $$

However, I see two issues with this reasoning:

1. The heuristic definition of $δ(x−a)$ as an "infinite spike at $x=a$" is informal. Since $δ(x−a)$ is not a classical function, we cannot speak about its value outside the integral.
2. The integral splitting above is valid for classical functions, but $\int_{-\infty}^{\infty} f(x) \delta(x-a)dx$ is not a standard Riemann or Lebesgue integral—it represents the action of a distribution.

So, is the property $\int_{a-\epsilon}^{a+\epsilon} f(x) \delta(x-a)dx=f(a)$ a rigorously proven result, or does it rely on an interpretive extension of the Dirac delta's sifting property?

Edit

Another approach might be to write:

$$\int_{a- \epsilon}^{a+\epsilon} f(x) \delta(x-a)dx=\int_{\mathbb{R}} f(x) 1_{(a-\epsilon,a+\epsilon)} (x)\delta(x-a)dx$$

as suggested in the comment. However, this approach, like the first one, involves treating $δ(x)$ outside the context of an integral over $\mathbb{R}$, which is the main issue. Really, our problem lies in the interval of integration, not the integrand. If such an interpretation were valid, the first approach would already have been sufficient.

Answer 1

Rigorously speaking (i.e. before using abuse of notations), there is no meaning to $\int_{a-\epsilon}^{a+\epsilon} f(x)\,\delta(x-a)\,\mathrm d x$ from the theory of distributions. In the theory of distributions, what has meaning is the action on a smooth test function $\varphi$, denoted for example $\langle \delta_a,\varphi\rangle$, which is by abuse of notations denoted by $\int_{\Bbb R} \varphi(x)\,\delta(x-a)\,\mathrm d x$.

But in the same way as distributions generalize functions, distributions also generalize measures, by writing if $\varphi\in C^\infty_c$ $$ \langle \mu,\varphi\rangle = \int_{\Bbb R} \varphi(x)\,\mu(\mathrm d x) $$ (the integral with respect to $\mu$), and any positive distribution has a unique associated measure (this is a Riesz representation-type theorem of Schwartz). In particular, in these cases, it allows to define $\langle\mu,\varphi\rangle$ for any $\varphi\in L^1(\mu)$ (integrable with respect to $\mu$).

In the case of the Dirac distibution, the associated measure (the Dirac measure) is defined on sets as $$ \delta_a(A) = \Bbb 1_{A}(a), $$ i.e. its value is $1$ if and only if $a\in A$, and $0$ else. Integration with respect to $\delta_a$ indeed yields $$ \int_{\Bbb R} \varphi(x) \,\delta_a(\mathrm d x) = \varphi(a). $$ and any function $\varphi$ continuous in $a$ is $\delta_a$-integrable. In particular taking $\varphi(x) = f(x)\,\Bbb 1_{[a-\epsilon,a+\epsilon]}(x)$ gives $$ \int_{a-\epsilon}^{a+\epsilon} \varphi(x) \,\delta_a(\mathrm d x) = \varphi(a). $$

Remarks:

• In some sense, the Dirac function could be first defined as a measure, and then it is in particular a distribution.
• If everything is not clear, I had written some more details for the link between distributions, functions and measures here Prove properties of Dirac delta from the definition as a distribution

Answer 2

On this forum there are many question by math students about the Dirac delta function, in the context of the field of distributions. Hence the subject is considered "higher mathematics" and therefore difficult. However, this is not necessary. Elementary mathematics is usually sufficient to understand the properties of the Dirac delta function $\delta (x-a)$.

First of all we start with an arbitrary function (a so-called nascent delta function) centered around the value $x = a$. We normalize the function, so that the surface area under the curve equals unity. There is a parameter $w$, which represents the width of the function. Now if we take the limit of $w$ to $+0$, so that the nascent delta function becomes a narrow spike, we arrive at the desired delta function.

In principle all choices for the nascent delta function are equivalent. So let us make a convenient choice, the block representation: $\delta (x-a) = 1/(2w)$ for $a-w \le x \le a+w$. We now consider the integral the OP found on Wolfram. We get a very simple integral, where two competing windows affect the integration boundaries. It is easy to see that $\epsilon$ will always end up being larger than $w$, because of the limit applied to the latter. Hence the $\epsilon$ window has no effect on the outcome. And therefore the result is the same as if the bounds were $-\infty$ and $+\infty$.