# I Lebesgue Integral of Dirac Delta "function"

1. Nov 17, 2017

### lightarrow

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

2. Nov 17, 2017

### WWGD

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..

3. Nov 17, 2017

### FactChecker

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.

4. Nov 17, 2017

### lightarrow

I reply to both.
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

5. Nov 17, 2017

### WWGD

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.

6. Nov 17, 2017

### FactChecker

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.

7. Nov 17, 2017

### WWGD

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.

8. Nov 17, 2017

### George Jones

Staff Emeritus
9. Nov 17, 2017

### PeroK

$+\infty \notin \mathbb{R}$

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

10. Nov 17, 2017

### WWGD

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.

11. Nov 17, 2017

### PeroK

That doesn't make $+\infty \in \mathbb{R}$.

12. Nov 17, 2017

### WWGD

True, but functions taking value in the extended Real numbers , i.e., with values in $[-\infty, \infty]$ are allowed.

13. Nov 17, 2017

### PeroK

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.

14. Nov 17, 2017

### WWGD

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.

15. Nov 17, 2017

### 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.

16. Nov 17, 2017

### 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.

17. Nov 18, 2017

### lightarrow

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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lightarrow

18. Nov 18, 2017

### PeroK

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 $2f$.

19. Nov 18, 2017

### FactChecker

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.

20. Nov 18, 2017

### StoneTemplePython

I think this is famously wrong. I'm pretty sure Pythagoras put someone to death over the irrationality of this solution.