Lebesgue-integrability of the Dirac delta function?

Question

I stumbled upon this question, which I think was answered incorrectly. Considering the Dirac delta function $$ \delta\colon\mathbb R\to[0,\infty],x\mapsto\begin{cases}\infty,&\text{if }x=0,\\0,&\text{otherwise}\end{cases} $$ as a mapping from $\mathbb R$ to $[0,\infty]$ (which is perfectly well-defined) and NOT as a distribution $\mathcal S(\mathbb R)\to\mathbb R$ or $C_c^\infty(\mathbb R)\to\mathbb R$. I commonly see people use the sequence $\delta_n:=n\cdot1_{[-\frac1{2n},\frac1{2n}]}$ to "prove" $\int\delta\,d\lambda=1$, but this sequence isn't even increasing. If this argument actually holds, what prevents me from using $\delta_n:=\alpha\cdot n\cdot1_{[-\frac1{2n},\frac1{2n}]},\alpha>0$ to "prove" $\int\delta\,d\lambda=\alpha$? Now my question is whether or not the function $\delta$ as defined above is actually Lebesgue-integrable? Here's my shot at proving that it is Lebesgue-integrable (maybe I just missed some detail):

$\delta\colon\mathbb R\to[0,\infty]$ is $\mathcal B(\mathbb R)$-$\mathcal B(\overline{\mathbb R})$-measurable, because we have $$\delta^{-1}(\mathcal B(\overline{\mathbb R}))=\{\emptyset,\{0\},\mathbb R\setminus\{0\},\mathbb R\}\subseteq\mathcal B(\mathbb R).$$ Furthermore $\delta_n:=n\cdot1_{\{0\}}\colon\mathbb R\to[0,\infty[$ is a simple function (non-negative, bounded, measurable and has finite image) for every $n\geq1$ and the sequence $(\delta_n)_{n\geq1}$ is monotonically increasing, converges pointwise to $\delta$ and it holds that $$\lim_{n\to\infty}\int\delta_nd\lambda=\lim_{n\to\infty}n\cdot\underbrace{\lambda(\{0\})}_{=0}=0<\infty.$$ Hence, $\delta$ is Lebesgue-integrable by definition and we have $\int\delta\,d\lambda=0$.

APPENDIX: Here is the definition of Lebesgue-integrability that I've learned. Let $(\Omega,\mathcal A,\mu)$ be some measure space. A function $f\colon\Omega\to[0,\infty)$ is called simple, if it is $\mathcal A$-$\mathcal B(\mathbb R)$-measurable and has a finite image, i.e. $f=\sum_{i=1}^k\alpha_i1_{A_i}$ for some $\alpha_1,\dots,\alpha_k\in[0,\infty[$ and $A_1,\dots,A_k\in\mathcal A$ (not necessarily disjoint). It's integral is defined as $\int f\,d\mu:=\sum_{i=1}^k\alpha_i\mu(A_i)\in[0,\infty]$. We then showed that this is well-defined, i.e. independent of the choice of $\alpha_1,\dots,\alpha_k$ and $A_1,\dots,A_k$. Now let $f\colon\Omega\to[0,\infty]$ be $\mathcal A$-$\mathcal B(\overline{\mathbb R})$-measurable. We proceeded to show that there exists an increasing sequence $(f_n)_{n\in\mathbb N}$ of simple functions $f_n\colon\Omega\to[0,\infty)$ such that $f_n\to f$ pointwise. We then defined the integral of $f$ to be $\int f\,d\mu:=\lim_{n\to\infty}\int f_nd\mu\in[0,\infty]$ and showed that this is well-defined too, i.e. independent of the choice of $(f_n)_{n\in\mathbb N}$. We called $f$ integrable, if $\int f\,d\mu<\infty$. A $\mathcal A$-$\mathcal B(\overline{\mathbb R})$-measurable function $f\colon\Omega\to\overline{\mathbb R}$ is called integrable, if $f^+:=f\cdot1_{\{f\geq0\}}\colon\Omega\to[0,\infty]$ and $f^-:=-f\cdot1_{\{f\leq0\}}\colon\Omega\to[0,\infty]$ are integrable. In this case we define $\int f\,d\mu:=\int f^+d\mu-\int f^-d\mu$.

Using this definition, it is easy to show using measure theoretic induction that a $\mathcal A$-$\mathcal B(\overline{\mathbb R})$-measurable function $f\colon\Omega\to\overline{\mathbb R}$ with $\mu(\{f\neq0\})=0$ is integrable and has integral 0.

$\delta\colon\mathbb R\to[0,\infty]$ clearly is $\mathcal B(\mathbb R)$-$\mathcal B(\overline{\mathbb R})$-measurable and satisfies $\lambda(\{\delta\neq0\})=\lambda(\{0\})=0$. Hence $\delta\colon\mathbb R\to[0,\infty]$ is integrable and $\int\delta\,d\lambda=0$.

Answer 1

It depends on your definition of Lebesgue Integrable...

If $\int \delta d \lambda := \lim_{n \to \infty} \int \delta_n d \lambda,$ then, since the limit exists and is 1, $\delta$ is integrable.

However, usually the Lebesgue integral is defined as $$\int \delta d \lambda := \sup\left\{ \sum_{i=1}^n (b_i-a_i)\alpha_i \right\},$$ where the $\sup$ runs over all functions $\varphi(t) := \sum_{i=1}^n \alpha_i \cdot 1_{(a_i,b_i)}(t),$ that satisfy $0 \leq \varphi(t) \leq \delta(t).$ (Here $1_S(t)$ is the indicator function of the set $S$.) In this definition, $\int \delta d \lambda = 0$, so $\delta$ is integrable for a stupid reason: it's 0 almost everywhere.

PS: A generally difficult problem in analysis is to identify when $\lim \int f_n = \int \lim f_n.$ With your $\delta$ and $\delta_n$ functions, you have an example of when the two sides are not equal.

Answer 2

Yes of course if you define a function $\delta:\Bbb R\to[0,\infty]$ that way then $\delta\in L^1$, because $\delta$ is measurable and $\int|\delta|=0<\infty$.

But this has no relevance whatever to the correctness of that other answwer! That other question is about the object commonly known as the "Dirac delta", and the function you defined here is not the Dirac delta. (This serves to answer the things you say you're confused about: In each case the confusion arises from assuming that that function is the Dirac delta.)