Integrability of sums of Dirac deltas

Question

this is my first post in the forum and I am an engineer, so I apologize in advance if my question is not clearly stated.

Consider the function

$f(x)=\sum_{i=1}^N a_i\delta(x-x_i)$ where $a_i\in\mathbb{R}$ for all $i$ and $\delta(\cdot)$ denotes the Dirac delta.

Is this function Lebesgue integrable? I am interested in both the case when $N$ is finite, and when $N$ is infinite. I have read a few posts in the forum where the Dirac delta is claimed not to be integrable. I would very much appreciate some help with this or some pointers.

Answer 1

Here's what the people on the forum meant:

Say $\lambda$ is the Lebesgue measure. There is no $\lambda$-measurable function $f(x-x_i)$ where:

• $\int_S f(x-x_i) d\lambda=1$ if $x_i \in S,$ and
• $\int_S f(x-x_i) d\lambda=0$ if $x_i \notin S,$ where $\int[\dots]$ is a Lebesgue integral.

This is due to the fact that the Lebesgue measure of a single point is 0: $\lambda(\{x_i\})=0.$

Those were the two properties you want your 'delta function' object to have. So you can infer that if you're using the Lebesgue integral, then the 'delta function' can't be a $\lambda$-measurable function.

What is done within the theory of the Lebesgue integral instead is to think of $\delta(x-x_i)$ as a measure, and integrate with respect to it. We can say that if $S$ is a (borel) set of $\mathbf{R}$, then if $x_i\in S$, $[\delta(x-x_i)](S)=1$, and otherwise $[\delta(x-x_i)](S)=0$.

Then if $f:\mathbf{R}\to \mathbf{R}$ is a (Borel-measurable) function , then the Lebesgue integral of $f$ over a (Borel) set $S$ with respect to $\delta(x-x_i)$ is: $\int_S f\ d[\delta(x-x_i)]=f(x_i).$

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Under this interpretation of $\delta$, then your sum is a sum of measures on $\mathbf{R},$ which may or may not define a new (possibly signed) measure. You're interested in 'the integral of the sum,' which translates to the measure of the whole space under your new measure, which can be re-written as a Lebesgue integral: $$\mu(\mathbf{R}) = \int_{\mathbf{R}}1 \ d\mu.$$

You wanting the series to be 'Lebesgue integrable' means that you want $\int_{\mathbf{R}}1 \ d\mu$ to be a finite-valued Lebesgue integral. This is true when

• $\mu$ is a finite signed measure, and
• $\mu({\mathbf{R}})<\infty$.

For both these criteria to hold, the set of $a_i$s must fall into one of the following categories:

• $N$ is finite. Then you can check that $\sum_{i=1}^N a_i$ a finite sum, so by construction $\mu$ is a finite signed measure and $\mu(\mathbf{R})=\sum_i a_i$.
• $N$ is countably infinite and $\sum_i |a_i|<\infty$. Then the series is absolutely convergent, so $\mu$ can be decomposed into the difference between two signed measures defined by: $\mu_+(S):=\sum_{i:x_i\in S} \max(a_i,0)$ and $\mu_-:=\sum_{i:x_i\in S}\max(-a_i,0)$ with $\mu(S)=\mu_+(S)-\mu_-(S)$. You can check that all those measures are finite, and $\mu(\mathbf{R})=\sum_{i=1}^\infty a_i$ is finite too.
• The indexing set is uncountably infinite, and all but a countable number of $a_i$ are zero, and among those that are nonzero, $\sum_i |a_i|<\infty$ (see the second bullet point).

These are the only cases in which your summation is 'Lebesgue integrable' in some sense.

If your series does not meet any of the above criteria, you still may be able to think of its 'integral' as 'finite' (e.g. if $\sum_i a_i$ diverges, but its sequence of partial sums converges, or converges in some regularization sense). But then it wouldn't be a Lebesgue integral, at least without some deep extensions to the theory.

One possible avenue for these cases is distribution theory.

Answer 2

For finite or a countable sum, the integral exists and is a.e. zero. The value of the integral of f(x) is $\sum(a_i)$ which comes directly from the definition of distributions.

$$\int f(x) dx = \int \sum_i a_i \delta(x - x_i) dx = \sum_i a_i \int \delta(x - x_i) dx = \sum_i a_i$$

Remember that $f(t) = \int(c(x) \delta(t - x) dx$

and that this is no different than your sum for suitable $c(x)$. e.g., if $c(x) = a_i\mbox{ when } x = x_i\mbox{ else } 0$.

The integrability of the dirac delta function is moot. It is defined and manipulated in such a way to work in certain cases. The question is, does the integral of your sum make sense and it does in the same sense as dirac does in the first place. (the point is, if anyone claims it doens't they are claiming dirac doesn't because this can be shown by the rules of distributions to be a valid distribution function. The question of Lebesgue measurability is rather moot too. $f(x) = 0$ almost everywhere and hence the Lebesgue integral is 0. So, it is Lebesgue measurable, but it won't give the same answer since it simply ignores all the interesting points. Hence, the appropriate context here is that of distributions, after all, that is what you defined your function to be in the first place.