# Lebesgue Integral of Dirac Delta "function"

• I
Is the "function" R->R
f(x) = +oo, if x =0 (*)
0, if x =/= 0
Lebesgue measureable? Does its Lebesgue Integral exist? If yes, how much is it?

(*) Certainly we shoud give a convenient meaning to that writing.

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lightarrow

WWGD
Gold Member
Is the "function" R->R
f(x) = +oo, if x =0 (*)
0, if x =/= 0
Lebesgue measureable? Does its Lebesgue Integral exist? If yes, how much is it?

(*) Certainly we shoud give a convenient meaning to that writing.

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lightarrow
In measure theory ( though not so in standard analysis/calculus) you get ##\infty*0=0 ##. But in most respects this is not seen as an actual function, since then you get the contradicting result that ##\int_{-\infty}^{\infty} \delta dx=1##. So you need to treat it as a functional or at least not as a function in order to avoid a contradiction, EDIT since, if assumed to be a function its integral would be 0..

FactChecker
Gold Member
The Lebesgue integral of the delta function would be zero, since it equals 0 almost everywhere and the Lebesgue integral of g(x)≡0 is 0.

There are rigorous treatments of the delta function. They require measure theory or the theory of distributions and test functions.

I know Dirac Delta is better treated as a distribution (I've studied them), but my question is related to what I wrote.
Has it any meaning to write a function the way I wrote it? If it has, in which sense should I consider the write "f(0) = +oo"? Then, is it Lebesgue measureable? WWGD answered to this writing that the integral is 0, because that "function" is different than 0 just in a set which has Lebesgue measure = 0. Since I'm not an expert of Lebesgue thery, my question is if that is possibile even for values of x where f = +oo. I only have seen definitions of "Lebesgue measureable function f in (a, b) " when f is bound inside the interval (a, b) (can be infinite at the limit for x->a or x->b).

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WWGD
Gold Member
I know Dirac Delta is better treated as a distribution (I've studied them), but my question is related to what I wrote.
Has it any meaning to write a function the way I wrote it? If it has, in which sense should I consider the write "f(0) = +oo"? Then, is it Lebesgue measureable? WWGD answered to this writing that the integral is 0, because that "function" is different than 0 just in a set which has Lebesgue measure = 0. Since I'm not an expert of Lebesgue thery, my question is if that is possibile even for values of x where f = +oo. I only have seen definitions of "Lebesgue measureable function f in (a, b) " when f is bound inside the interval (a, b) (can be infinite at the limit for x->a or x->b).

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lightarrow
Hi, yes, the value ##\infty## may be assumed by functions in the case of Lebesgue integration. And in that case the algebra is that ##\infty \times 0 =0 ##. This is not true in a more general sense, where you must be using limits, etc.

FactChecker
Gold Member
It is Lebesgue measurable. It is equal to 0 almost everywhere. A better question is "is it defined". Its definition δ(0)=∞ is not considered valid in standard real analysis.

• WWGD
WWGD
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It is Lebesgue measurable. It is equal to 0 almost everywhere. A better question is "is it defined". Its definition δ(0)=∞ is not considered valid in standard real analysis.
You may even formally do a Lebesgue sum, in the same sense that you can show a Riemann integral has a value by doing a Riemann sum.

PeroK
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Is the "function" R->R
f(x) = +oo, if x =0 (*)
0, if x =/= 0
Lebesgue measureable? Does its Lebesgue Integral exist? If yes, how much is it?

(*) Certainly we shoud give a convenient meaning to that writing.

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lightarrow
##+\infty \notin \mathbb{R}##

Therefore, this is not a real-valued function and not well defined.

WWGD
Gold Member
##+\infty \notin \mathbb{R}##

Therefore, this is not a real-valued function and not well defined.
Lebesgue integration allows for infinite-valued functions, though they may take the value infinity in a set of measure zero in order for the Lebesgue integral to exist.

PeroK
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Lebesgue integration allows for infinite-valued functions, though they may take the value infinity in a set of measure zero in order for the Lebesgue integral to exist.
That doesn't make ##+\infty \in \mathbb{R}##.

WWGD
Gold Member
That doesn't make ##+\infty \in \mathbb{R}##.
True, but functions taking value in the extended Real numbers , i.e., with values in ##[-\infty, \infty] ## are allowed.

PeroK
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True, but functions taking value in the extended Real numbers , i.e., with values in ##[-\infty, \infty] ## are allowed.
If I let ##f: \mathbb{R} \rightarrow \mathbb{R}##, such that:

##f(x) = \sqrt{x}##

Then that function is not well-defined. It makes no difference, for example, that you can integrate a complex-valued function.

WWGD
Gold Member
If I let ##f: \mathbb{R} \rightarrow \mathbb{R}##, such that:

##f(x) = \sqrt{x}##

Then that function is not well-defined. It makes no difference, for example, that you can integrate a complex-valued function.
What I meant is that the Lebesgue integral of a function taking values in ##[-\infty, \infty] ## is well-defined.EDIT: meaning that the question of whether the integral exists may be answered as yes or no.

Svein
The "Dirac delta" is not a function. The best way to look at it is as a Stieltjes integral where the point {0} has measure 1 and the measure of any set that does not contain {0} is 0.

The Stieltjes integral: https://en.wikipedia.org/wiki/Riemann–Stieltjes_integral allows singular points to have a positive measure.

mathman
The major (?) difficulty with Dirac delta function is that it is useful to physicists, who don't have to content with mathematical niceties. Mathematicians have to use either distributions or Stieltjes integrals to treat it rigorously.

Thank to all for the answers.
My main doubt is the following.
For what (little) I know about lebesgue integral, we make a partition y_i of the set of all the possible values taken from the function, then we compute, for every interval (y_i; y_{i+1}) , the measure mu_i of the corresponding set of values of x, we choose a value Y_i inside that interval of y which is representative of them, then we compute Sum_i mu_i*Y_i.
Is this correct?
1. The partition of the values y have to be finite?
2. If is possible to consider +oo as "representative value of some interval (y_s; y{s+1})" and noticed that mu_s = 0, how can I deduce (as WWGD wrote at the beginning) that +oo*0 = 0? It seems meaningless to me. The only "reasonable" way to give a meaning to the lebesgue integral in this case *given that the integral exist at all* is to consider functions f_n the graph of which become even more "short" and tall for n->oo and the area under of it is always 1, as when me manipulate the Dirac Delta as distribution.
Or that such area is k, or +oo, or 0, depending on how we want to define our "function" in general. In conclusion, I can't understand why +oo*0 have to be 0 here.

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PeroK
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What I meant is that the Lebesgue integral of a function taking values in ##[-\infty, \infty] ## is well-defined.EDIT: meaning that the question of whether the integral exists may be answered as yes or no.
You're missing the point that mathematics doesn't work like that. The equation:

##x^2 = 2## has no integer solutions, but does have a rational solution.

Whether it has a solution depends on the set of numbers you are dealing with.

Similarly, ##f: \mathbb{R} \rightarrow \mathbb{R}## such that ##f(0) = +\infty## is not a well-defined function, and therefore cannot be integrated. The function ##f: \mathbb{R} \rightarrow [-\infty, +\infty]## such that ##f(0) = +\infty## is well-defined and can be integrated, assuming your integral is defined for such functions.

As mentioned in some posts above, a physicist doesn't have to worry about such niceties, but a mathematician does. And, since this is in a maths forum, only well-defined functions please!

(Ironically, if the OP had omitted the domain and range, we could have implicitly accepted the range of the function. But, the OP is quite explicit about trying to define a real-valued function.)

In conclusion, I can't understand why +oo*0 have to be 0 here
Mathematically, you are free to define ##\infty.0## to be whatever you want it to be.

One problem with taking this product to be ##1##, say, is what happens to the integral of the function ##2f##.

• FactChecker
FactChecker
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I think it's hard to tell the OP that what he wants to do is invalid, when there are mathematical approaches that put similar things on a rigorous, valid basis. The OP should be aware that a mathematical approach like generalized functions (see https://en.wikipedia.org/wiki/Generalized_function ) has been developed to achieve what he wants to do with complete mathematical rigor. And Lebesgue integration was an important step in that direction. The rigorous approach does a lot to identify what is valid to do and where the treacherous issues are that have counterexamples.

StoneTemplePython
Gold Member
The equation:

##x^2 = 2## has no integer solutions, but does have a rational solution.
I think this is famously wrong. I'm pretty sure Pythagoras put someone to death over the irrationality of this solution.

• rude man and PeroK
PeroK
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I think this is famously wrong. I'm pretty sure Pythagoras put someone to death over the irrationality of this solution.
Ha! That serves me right for being too pedantic!

I think it's hard to tell the OP that what he wants to do is invalid, when there are mathematical approaches that put similar things on a rigorous, valid basis. The OP should be aware that a mathematical approach like generalized functions (see https://en.wikipedia.org/wiki/Generalized_function ) has been developed to achieve what he wants to do with complete mathematical rigor. And Lebesgue integration was an important step in that direction. The rigorous approach does a lot to identify what is valid to do and where the treacherous issues are that have counterexamples.
Ok but I would accept the fact the Lebesgue Integral in this case /doesn't exist/, not that it exists and is 0.

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Wakinian Tanka

FactChecker
Gold Member
Ok but I would accept the fact the Lebesgue Integral in this case /doesn't exist/, not that it exists and is 0.
I am not an expert at this, but here are my two cents:
The delta "function" presents difficulties if you try to treat it as a normal function. But notice that it is really defined by the properties of its integral. You can get the same integration results by defining it as a "generalized function" or as a "measure". Generalized functions are split into the smooth part and the "singular part". Multiplication rules are not the same as for functions. As long as they are within integrals, they can be dealt with.

PS. If you are to do much with Lebesgue integrals, you will get used to the fact that the value of the integral does not depend on value of the integrand on a set of measure zero.

PPS. I think this was a great series of lectures that use delta functions extensively ( ). He teaches the proper use of the delta function without getting too tied up in advanced mathematical proofs.

Last edited:
WWGD
Gold Member
You're missing the point that mathematics doesn't work like that. The equation:

##x^2 = 2## has no integer solutions, but does have a rational solution.

Whether it has a solution depends on the set of numbers you are dealing with.

Similarly, ##f: \mathbb{R} \rightarrow \mathbb{R}## such that ##f(0) = +\infty## is not a well-defined function, and therefore cannot be integrated. The function ##f: \mathbb{R} \rightarrow [-\infty, +\infty]## such that ##f(0) = +\infty## is well-defined and can be integrated, assuming your integral is defined for such functions.

As mentioned in some posts above, a physicist doesn't have to worry about such niceties, but a mathematician does. And, since this is in a maths forum, only well-defined functions please!

(Ironically, if the OP had omitted the domain and range, we could have implicitly accepted the range of the function. But, the OP is quite explicit about trying to define a real-valued function.)

Mathematically, you are free to define ##\infty.0## to be whatever you want it to be.

One problem with taking this product to be ##1##, say, is what happens to the integral of the function ##2f##.
Ok, I did not get where you were going, but we abandoned all Mathematical niceties when we called it a function in the standard sense; this assumption leads to a cntradiction -- a function with support of measure 0 having integral equal to 1 -- so we left that port long ago.

WWGD
Gold Member
You're missing the point that mathematics doesn't work like that. The equation:

##x^2 = 2## has no integer solutions, but does have a rational solution.

Whether it has a solution depends on the set of numbers you are dealing with.

Similarly, ##f: \mathbb{R} \rightarrow \mathbb{R}## such that ##f(0) = +\infty## is not a well-defined function, and therefore cannot be integrated. The function ##f: \mathbb{R} \rightarrow [-\infty, +\infty]## such that ##f(0) = +\infty## is well-defined and can be integrated, assuming your integral is defined for such functions.

As mentioned in some posts above, a physicist doesn't have to worry about such niceties, but a mathematician does. And, since this is in a maths forum, only well-defined functions please!

(Ironically, if the OP had omitted the domain and range, we could have implicitly accepted the range of the function. But, the OP is quite explicit about trying to define a real-valued function.)

Mathematically, you are free to define ##\infty.0## to be whatever you want it to be.

One problem with taking this product to be ##1##, say, is what happens to the integral of the function ##2f##.
Yes, I get it, I am a Mathematician myself, yet it seems all it takes is extending the codomain ; it does not seem like so serious of an issue. The more serious issue is that assuming it is a function leads to a contradiction.