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 $\int\delta\,d\lambda=0$.
Best Answer
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.