1
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ $\lim_{n \to \infty} n 1_{[-1/(2n),1/(2n)]}$ is the definition of the Dirac delta distribution. This sequence converges in the sense of distributions because for any $\phi \in C^\infty_c$ then $\lim_{n \to \infty} \phi \ast n 1_{[-1/(2n),1/(2n)]} \in C^\infty$. As you see $\begin{cases}\infty,&\text{if }x=0,\\0,&\text{otherwise}\end{cases}$ doesn't define any distribution uniquely. $\endgroup$
    – reuns
    Feb 10, 2019 at 4:44
  • $\begingroup$ Okay and where does the followong argument break down? Every measurable function that vanishes almost everywhere is Lebesgue-integrable and has integral 0. $\delta$ as defined in my question is measurable and vanishes almost everywhere. Hence, it is Lebesgue-integrable and has integral 0. I can find at least 34 more theorems and propositoons in my Analysis III script (yeah I actually counted them) that imply the integrability of $\delta$ as defined in my question and that it has integral 0. $\endgroup$
    – Cubi73
    Feb 10, 2019 at 10:31
  • $\begingroup$ Every "measurable function" : $\delta$ isn't a function, it is a distribution. Removing finitely many (or countably many) points of a function doesn't change its $\int$, doing so with $\delta$ does change it, that's the difference. $\endgroup$
    – reuns
    Feb 10, 2019 at 21:22
  • $\begingroup$ @reuns Then take a closer look at how I defined $\delta$. I explicitly stated that I consider the function above mapping $\mathbb R$ to $[0,\infty]$ and not the distribution. $\endgroup$
    – Cubi73
    Feb 10, 2019 at 21:54
  • 1
    $\begingroup$ It is $= 0$ in $L^1$. For Lebesgue integral we are just looking at the measures of sets of the form $\{ x, f(x) \ge M\}$. Here $\{ x, f(x) \ge M\} = \{0\}$ has measure $0$ $\endgroup$
    – reuns
    Feb 10, 2019 at 21:57

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ Why does the limit definition yield $\int\delta\,d\lambda=1$? $\endgroup$
    – Cubi73
    Feb 10, 2019 at 2:49
  • $\begingroup$ The $n$-th integral in the limit is 1, and $\lim_{n \to \infty} 1 = 1.$ $\endgroup$
    – Dzoooks
    Feb 10, 2019 at 2:50
  • $\begingroup$ Okay, let me ask this differently: Where does my proof outlined in my question break down? I used the following definition of Lebesgue-integrability: A non-negative function $f\colon\mathbb R\to[0,\infty]$ is called Lebesgue-integrable, if there exists an increasing sequence $(f_n)_{n\in\mathbb N}$ of simple functions $f_n\colon\mathbb R\to[0,\infty[$ such that $(f_n)_{n\in\mathbb N}$ converges pointwise to $f$ and the sequence $(\int f_nd\lambda)_{n\in\mathbb N}$ is bounded. In this case we call $\int f\,d\lambda:=\lim_{n\to\infty}\int f_nd\lambda$ the integral of $f$. $\endgroup$
    – Cubi73
    Feb 10, 2019 at 2:59
  • $\begingroup$ Just to clarify, I didn't use Lebesgue's Theorem to justify the limit, it is part of the definition of Lebesgue-integrability that I know. $\endgroup$
    – Cubi73
    Feb 10, 2019 at 3:03
  • $\begingroup$ @Cubinator73 Oh, you are asking us to critique your proof? The question was unclear. What you have is just about accurate, except you should really show that ANY sequence $f_n$ satisfying $0 \leq f_n(x) \leq \delta(x)$ has $\lim_n \int f_n = 0$. Not just $\delta_n$. But it's pretty much the same argument... $\endgroup$
    – Dzoooks
    Feb 10, 2019 at 3:06
1
$\begingroup$

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.)

$\endgroup$
4
  • $\begingroup$ Well, I tried to distinguish between them by calling them Dirac delta function and Dirac delta distribution and in addition to that I explicitly stated that I consider the function defined in my question and not the distribution. Though, you are right, I actually thought that the $\delta$ I defined is called Dirac delta function. $\endgroup$
    – Cubi73
    Feb 10, 2019 at 14:39
  • $\begingroup$ @Cubinator73 Fine. The two are not the same - that other question is about the Dirac delta distribution, so nothing you say about the Dirac delta function has any relevance to that question. $\endgroup$ Feb 10, 2019 at 14:41
  • $\begingroup$ Okay, just one last thing. The Dirac delta distribution $\delta\colon C_c^\infty(\mathbb R^d)\to\mathbb R,\varphi\mapsto\varphi(0)$ is not regular, meaning there is no $f\in L_{\text{loc}}^1(\mathbb R^d)$ such that $\delta(\varphi)=\int f\varphi d\lambda^d$ for every $\varphi\in C_c^\infty(\mathbb R^d)$. Am I right in assuming we still like to write $\delta(\varphi)=\int\varphi f\,d\lambda^d=\varphi(0)$ even though there is no such $f$ that satisfies this property and thus make it a convention? $\endgroup$
    – Cubi73
    Feb 10, 2019 at 14:52
  • 1
    $\begingroup$ Do "we" like to write it that way? Some people do; see millions of posts here that write $\int \phi(t)\delta(t)\,dt$, or various differential equations books. Mathematicians don't - it leads to confusion. $\endgroup$ Feb 10, 2019 at 16:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .