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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.

Michael Hardy
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OSE
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    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. – Daniel Fischer Oct 11 '13 at 14:19
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    "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... – Tomás Oct 11 '13 at 14:27
  • Note that, its integral on a set of measure zero is $1$!. – Mhenni Benghorbal Oct 11 '13 at 14:29
  • To me the fact that it's not a function in the sense of a mapping that takes an argument to a value seems to obvious for anyone to be confused about. I think the notation that "pretends it's a function" should have the same status as notation that "pretends" that complex numbers are numbers. – Michael Hardy Oct 11 '13 at 16:25
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    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..... – Michael Hardy Oct 11 '13 at 16:31
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    ....having seen that introductory material. – Michael Hardy Oct 11 '13 at 16:32
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    @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. – OSE Oct 11 '13 at 17:18
  • @MichaelHardy: for the sake of correcteness, what you describe is not a measurable function. A measurable function is one such that inverse images of open sets are measurable. To do integration, you don't need measurability in the codomain, but you need a topology. – Martin Argerami Mar 30 '18 at 19:39
  • @MartinArgerami : ok, Let's say inverse-images of Borel sets are measurable, and that follows from inverse-images of open sets being measurable, which therefore is sufficient. And "Borel" smells a whole lot like "measurable", but it's not exactly synonymous. – Michael Hardy Mar 30 '18 at 22:49
  • I'm not sure about the "smell" but yes, you have a point :) – Martin Argerami Mar 30 '18 at 23:29

1 Answers1

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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.

KCd
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Тимофей Ломоносов
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  • 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? – OSE Oct 11 '13 at 14:33
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    "Considering this a function leads" also to me living at the bottom of the ocean with my friends the Mermaids. – Did Oct 11 '13 at 16:42
  • @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. – Тимофей Ломоносов Oct 11 '13 at 17:23