The Riemann-Lebesgue lemma says that the Fourier transform of any $L^1$ integrable function on $\mathbb{R}^{d}$ satisfies:

$$\hat{f}(z):=\int_{\mathbb{R}^{d}} f(x) e^{-iz \cdot x}\,dx \rightarrow 0\text{ as } |z|\rightarrow \infty$$

This does not seem to be the case if $f(x) = \delta(x)$ which leads me to believe that the Dirac delta function is not $L^1$ integrable.

However, the Dirac delta function satisfies the following relation:

$$\int_{-\infty}^\infty f(x) \, \delta\{dx\} = f(0).$$

This seems to me that it is $L^1$ integrable.

Note: Please keep in mind that I come from an engineering background so I am mostly familiar with Riemann integration and have very little understanding of Lebesgue integration.

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    $\begingroup$ It's not even a function. It's a distribution, more specifically, it's a (positive) measure. Notation that pretends it were a function is an abomination that only leads to confusion. $\endgroup$ Commented Oct 11, 2013 at 14:19
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    $\begingroup$ "Notation that pretends it were a function is an abomination that only leads to confusion". I do agree with you @DanielFischer, this notation confused me so much... $\endgroup$
    – Tomás
    Commented Oct 11, 2013 at 14:27
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    $\begingroup$ I wonder how $L^1$ and $\delta$ get introduced in courses other than math courses. In math courses one is told that $L^1$ is the set of all measurable functions $f$ for which $\int|f|<\infty$, and measurable functions are mappings taking arguments to values, that have the property that inverse images of measurable sets under such functions are measurable. The delta function gets introduced in a way that makes it clear that it's quite a different thing---not even a "function" in the sense in which that term is most often used. Some people must be reading about these things without..... $\endgroup$ Commented Oct 11, 2013 at 16:31
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    $\begingroup$ ....having seen that introductory material. $\endgroup$ Commented Oct 11, 2013 at 16:32
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    $\begingroup$ @MichaelHardy This question is related to my research in fluid dynamics. Unfortunately the math courses for engineers are quite watered-down and my research often exposes me to topics for which I haven't seen the introductory material. $\endgroup$
    – OSE
    Commented Oct 11, 2013 at 17:18

1 Answer 1


As Daniel Fisher has already stated, it's not even a function but some functional.

Let's suppose for sometime that it is a function. It's easy to see that she's almost everywhere equal to $0$, the set $S:\{x:\delta(x)\ne0\}=\{0\}$ is countable and hence has Lebesgue measure of $0$.

If the function were integrable, one would have $\int\limits_{-\infty}^{+\infty}\delta(x) \, dx=0$.

But the function in question does have the following property: $\int\limits_{-\infty}^{+\infty}\delta(x) \, dx=1$ that contradicts with its supposed integrability.

Hence considering this a function leads to it being unintegrable by Lebesgue.

  • $\begingroup$ Thanks for the answer, was I correct in assuming that the Dirac delta does not satisfy the Riemann-Lebesgue lemma since it is unintegrable by Lebesgue? $\endgroup$
    – OSE
    Commented Oct 11, 2013 at 14:33
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    $\begingroup$ "Considering this a function leads" also to me living at the bottom of the ocean with my friends the Mermaids. $\endgroup$
    – Did
    Commented Oct 11, 2013 at 16:42
  • $\begingroup$ @OSE, when you consider it a function yes, but as it has been twice stated this is no function. Classicaly it is some linear bounded functional but I tried writing the answer in such a way that you could easily understand it. $\endgroup$ Commented Oct 11, 2013 at 17:23

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