Lebesgue integral of Dirac delta If I recall correctly, for a bounded function $f$
$$ \int_{\mathbb{R}} f \, d\mu = \int_{\mathbb{R} \setminus \{ a \} } f \, d\mu + f(a) \mu (a).$$
For the Lebesgue measure, $\mu(a) = 0$ and
$$ \int_{\mathbb{R}} f \, d\mu = \int_{\mathbb{R} \setminus \{ a \} } f \, d\mu.$$
But what about distributions that are not bounded? For example, the Dirac delta $\delta(x-a)$? Here $f(a) = \delta(0) = \infty$ (formally) and $\mu(a) = 0$ so $f(a) \mu (a)$ is indeterminate.
 A: Here is how you can make sense of: $\int _{-\infty}^{x}\delta(t)dt$.
$\int_{-\infty}^{x}\delta(t)dt = \int_{-\infty}^{\infty}\delta(t)H(x-t)dt = H*\delta = H(x)$. Here $*$ represents the operation of convolution. Note that the convolution of two distributions is justified as long as one of the distributions is compactly supported, see e.g. $\it{Functional Analysis, 2nd Ed., Rudin}$.
Hence, $\int_{\mathbb{R}}\delta(t)dt = 1$. Here is also a link to similar question that I asked recently : "Line integral" of delta distribution
A: Your formulas are correct if $f$ is measurable and integrable (or measurable and positive).
The delta function is not an actual function.
If you try to represent it as a function, it should vanish outside the origin.
But the origin has measure zero so it can be neglected as you justified.
You have just managed to prove that there is no $L^1$ function with the properties the delta function should have.
The tools of Lebesgue integration do not work for distributions (other than measurable functions) — other tools are needed to treat distributions.
Distributions are defined as functionals on some test function space.
You can formally represent them as integrating against "something that behaves like a function" (like Dirac's delta) and the usual functions are distributions in just this way.
Not all distributions are functions that could be treated like functions can, they have to be treated via duality to the space of test functions.
