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.

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
