Is the Dirac delta function $L^1$ integrable?

Question

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.

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.