# Dirac delta integral

Let $(X,\mathcal{A},\mu)$ be a measure space, and let $A \in \mathcal{A}$ be such that $\mu(A) = 0$. Then define $h\colon X \to [-\infty,\infty]$ by $h(x) = +\infty$ if $x \in A$ and $h(x) = 0$ otherwise. It's easy to see that $\int h d\mu = 0$ since, for example, one can take the sequence $f_n = n \chi_A$, all of which are simple, measurable, increasing and non-negative, and $f_n \to h$, so $\int h d\mu = \lim_n \int f_n d\mu = 0$. But our $h$ function is not so different from the Dirac delta, is it? If we take $(X,\mathcal{A},\mu) = (\mathbf{R},\mathcal{B},\lambda)$ (measure space of Borel sets equipped with the Lebesgue measure) then define $h$ using $A = \{0\}$ then we get the dirac delta. But, according to wikipedia, the integral of the Dirac delta (over $\mathbf{R}$) is $1$, contrary to our result with $h$.

If someone could explain what's going on and what I'm misunderstanding, I would be grateful. Thanks.

• – Don Larynx Nov 30 '13 at 20:12
• – Did Nov 30 '13 at 20:18

The Dirac delta is not a function, not even a function that takes the value $\infty$ at one point. In particular, it is not the limit (in any reasonable topology that I can think of) of the functions $f_n$ that take the value $n$ at $0$ and the value $0$ everywhere else.

If one wants to approximate delta by genuine functions, one needs to use functions whose integrals converge to $1$, and the convergence of the functions to delta will be in the sense of distributions, not pointwise convergence.

If $\mu(A) = 0$, then defining $h$ as you have, notice that $h = 0$ a.e.-$\mu$. This is also what you're finding by trying to define the Dirac delta function in this way. You ask a good question in that this is the motivation behind defining distributions as "generalized functions" since the Dirac delta function measurable with respect to $\mu$.

To this end, there are two common ways to introduce to the Dirac delta to realize the Dirac function as a true mathematical object:

The first is that we consider it a distribution, that is, it is a linear functional on smooth, compactly supported functions (test functions). In this definition we define $\delta_x(f) := f(x).$ Common abused notation here is that if $T$ is any distribution, then you write $T(f) =: \int f\, T$ which matches the intuitive Dirac function you want to develop $\int f\, \delta_x = f(x)$.

The second is to define the Dirac (probability) measure where on some measurable set $(\Omega,\mathcal{B})$ (where each singleton set is a measurable set), for $x \in \Omega$, you define the measure $\delta_x$ to be the defined by $$\delta_x(A) = \begin{cases} 1 & x \in A \\ 0 & \text{ otherwise} \end{cases}$$ You can now check that if $f$ is some (real/complex valued) measurable map on $\Omega$ then $\int f(\omega) \delta_x(d\omega) = f(x)$.

As far as trying to define the Dirac delta as the limit of functions, suppose that $f_n \geq 0$ are all measurable such that $f_n(y) \to 0$ for every $y \neq x$, $f_n(x) \to \infty$, and $\int f_n(y) dy = 1$. Intuitively $f_n \to \delta_x$ pointwise. However, these approximations to the Dirac delta are a good example of a pointwise convergence which does not yield an $L^1$ convergence.