"Line integral" of delta distribution

Question

I have seen engineers write something to the effect : $\int_{-\infty}^x \delta_{0}(t) dt = H(x)$. Here $\delta_0(x)$ is the dirac delta distribution concentrated at origin and $H(x)$ is the step function. Even though the distributional derivative of $H(x)$ is the delta distribution, taking an integral of the delta distribution seems morally wrong, especially since $\delta(x)$ is not a lebesgue measurable function even. Besides an "intuitive" justification for this, is there a rigorous way to justify this?

Answer 1

Here is my attempt to justify this rigorously: Note that $\int_{-\infty}^{x}\delta(t)dt = \int_{-\infty}^{\infty}\delta(t)H(x-t)dt = H*\delta = H(x)$. 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. $\it{Functional Analysis, 2nd Ed., Rudin}$.

Answer 2

I tend to disagree with the approach that has been proposed.

In this case, I think that it would be better (or simpler, if you prefer) to think of $\delta_0$ simply as a measure, instead of a distribution. Namely, you have $$ \int_{\mathbb R} \chi_{(-\infty,x]}d\delta_0=H(x), $$ where the integral is the Lebesgue integral of the function $\chi_{(-\infty,x]}$ with respect to the measure $\delta_0$.

As you see there is no need for convolutions or really distribution theory altogether. Note also that $\delta_0$ is a Radon measure and so lives in the subset $(C^0_c)'$ of $(C^\infty_c)'$, no need for $C^\infty$ functions.