Given the Heaviside function, $$H := \begin{cases} 1 \quad \, x\geq0 \\ 0 \quad \, x < 0 \end{cases}$$
We can induce a measure that behaves in such a way so that all subsets of $A \subset \mathbb{R}$ are measurable and have non-zero measure iff $0 \in A$. The measure induced coming from the normal measure induced from monotonically increasing functions.
Now, I have two questions. When we integrate any function $f \in \mathscr{L}(\mu_H)$ w.r.t. $\mu_H$, we get
$$\int_\mathbb{R} f \, \mu_H = f(0).$$
The first thing I need to do is to show that there does not exist a function $g$ such that (for borel measurable subsets) $$\mu_H (A) = \int_A g \, dm$$ where $m$ is the lebesgue measure. Clearly no. For a function to have non-zero Lebesgue Integral w.r.t. the Lebesgue measure, it must be non-zero on a set with non-zero Lebesgue measure. So, such a $g$ would have to satisfy $$\mu_\alpha (A) = \int_A g \, dm >0 $$ if $0 \in A$ and $0$ otherwise. This is impossible, since $m(\{0\}) =0$, so if $$\int_A g \, dm > 0,$$ then $$\int_{A} g \, dm = \int_{A \setminus \{0\}} g \, dm > 0,$$
Now my other question is about the fact that there seems to be something going here that I can't quite put my finger on. I know the the $\delta$ distribution is the distributional derivative of the Heaviside function, and that the integral against $\delta_0$ picks out the function's value at $0$. So, why is it that when I integrate a function w.r.t. the measure induced by the heaviside function, it's as if I'm integrating against its distributional derivative? That is,
$$\int_\mathbb{R} f \, \mu_H = f(0) = \int_\mathbb{R} f H' \, dm$$
Is there a general fact here?