I Lebesgue Integral of Dirac Delta "function"

[WWGD]

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.

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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 $f(0)= +\infty$

 

[lightarrow]

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 $f(0)= +\infty$

Ok, I can also take $f_n(x) = (n/\sqrt(\pi)) exp[-(nx)^2]$

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[WWGD]

Ok, I can also take $f_n(x) = (n/\sqrt(\pi)) exp[-(nx)^2]$

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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.

 

[lightarrow]

Thanks to every one for the answers.

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