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.

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.

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.

Yes, it does exist ( while you need to correct the codomain of your function) and it equals 0 . I think the Lebesgue sum will show you this, as , in Lebesgue theory we define ## a \cdot \inft=0 ## for finite ##a##. The convergence criterion for Lebesgue integrals show it exists, and it equals ##0##. Only one of your partitions will contain a non-zero value. The problem is this contradicts the fact that the integral is supposed to equal ##1##.

Here goes: Start by defining [itex] F(x)=\int_{-\infty}^{x}\delta (t)dt[/itex]. Then F(x)=0 for x<0 and F(x)=1 for x≥0. It is therefore tempting to say that [itex]\delta (x)=dF(x) [/itex], except for the fact that neither side of the expression has any meaning at x=0. But for any interval [a, b] which does not contain {0}, the integral [itex] \int_{a}^{b}f(x)dF(x)=0[/itex] for any integrable function f(x). But going back to the definition of F(x), the integral [itex] \int_{-\varepsilon}^{\varepsilon}dF(x)[/itex] should be finite and equal to 1 for any ε>0.

The Stieltjes solution was to define dF(x) as a measure and it extends the concept of a measure as we know it from Lebesgue integration theory. In this case the measure of the point {0} is 1 and the measure of any interval that does not contain {0} is 0. Observe that the Stieltjes measure concept is much wider than the Dirac delta "function", as it contains all the standard interval measures. In addition, more than one singular point can have a measure different from 0.

Thanks for the link, I'll certainly will watch it. My initial question has been motivated from the fact a physics teacher stated that the Dirac Delta distribution is the only correct way of doing the integral, from - oo to +oo, of:

f(x) = 0, if x is different fro 0
f(x) = +oo if x =0

(and up to here I certainly agree with him)

/because the lebesgue integral of f(x) is zero/.

I replied to him that f(0) = +oo is meaningless (even in the codomain - oo +oo) and it's this the /first/ reason why the lebesgue integral of that f(x) cannot give the right answer.
If I "approximate" f(x) with the functions

f_n(x) = k*n for -1/n < x < 1/n
f_n(x) = 0 else

I get k as integral.
If I approximate f(x) with

g_n(x) = n^2 for -1/n < x < 1/n
g_n(x) = 0 else
I get +oo.

If I use:
h_n(x) = sqrt(n) for -1/n < x < 1/n
h_n(x) = 0 else
I get 0.

So, to me, it's impossibile to do the lebesgue integral of f(x) without precisly specifing what "f(0) = +oo" means. Or, the Lebesgue integral simply cannot be computed.

Oh, I agree with you there. I was careless when I said that the value on a set of measure zero didn't matter. It is not proper to include points where the function is not properly defined.

Regarding the video series, it is a long series (over 30), but I enjoyed it so much that I practically binge-watched it. I may watch it again.

Still, this is better and more rigorous than what I did , but you still do not get an actual value of 0 unless you extend continuously. You can also redefine , e.g., ##f(x)=1/x## so that ##f(0)= +\infty##

Sorry, don't see it here, ##f_n(0)=\frac {n}{\sqrt{\pi}} ##. My point is that , you can use sequential continuity in order to define ##f_n(0)##, and do not define it explicitly; just a (relatively minor) point.