Why is integral of differential delta function equal to delta function? Why is integral of differential delta function equal to delta function?
I.e. as I've read elsewhere:
$$\int_A 1 d \delta_A=\int_A d \delta_A=\delta_A=\delta_x(A)$$
https://planetmath.org/diracmeasure
Intuitively one could compare it to $\int 1 dx=x$, but since $\delta_A$ is a piece-wise function ...
Also I understand the intuition as to why the integral "ought to" equal that. But not the reasoning of "integral of differential equals variable of differential".
 A: Note that here $\delta_x$ is a measure, i.e. a function taking a set (belonging to some $\sigma$-algebra) and giving a nonnegative value. In this case, for any $A \subseteq \mathbb{R}$,
$$
\delta_x(A) := \begin{cases}
1 & \text{if } x\in A \\
0 & \text{if } x\not\in A
\end{cases}
$$
I won't go into the definitions of integrals of functions w.r.t. a measure, but for the measure $\delta_x$ we have
$$
\int f \, d\delta_x = f(x).
$$
Here, the $d$ in front of $\delta_x$ doesn't mean differential, but is rather just a common notation for telling with respect to what measure the integral is done. Some mathematicians would instead write $\int f(t) \, \delta_x(dt).$
An integral restricted to a set $A$ is defined by 
$
\int_A f \, d\mu = \int \chi_A f \, d\mu,
$
where $f$ is a function, $\mu$ a measure, and $\chi_A$ is the indicator function of $A$, defined by
$$
\chi_A(x) = \begin{cases}
1 & \text{if } x\in A \\
0 & \text{if } x\not\in A
\end{cases}
$$
Therefore,
$$
\int_A d\delta_x
= \int \chi_A \, d\delta_x
= \chi_A(x)
= \delta_x(A)
.
$$
