I have to evaluate the Lebesgue integral
$$ I = \int\limits_{[-1, 1]} \chi(x) \chi(x - \frac{1}{2}) d\left(\chi(x)\chi(x + \frac{1}{2})\right) $$
where $ \chi $ is the Heaviside function: $$ \chi(x) = \begin{cases}0, & x < 0 \\ 1, & x \geq 0 \end{cases} $$
The solution given is $$ I = \int\limits_{[-1, 1]} \chi(x) \chi(x - \frac{1}{2})d\left(\chi(x)\right) = \chi(0)\chi(-\frac{1}{2}) = 0 $$
but I'm unable to grasp it. Specifically, I don't understand what happened to $d\left(\chi(x)\chi(x + \frac{1}{2})\right)$.
Can you help me understand it?
Working in $[-1,1]$, we have $\chi(x) \chi(x - \frac{1}{2}) =\chi_{[\frac{1}{2},1]}(x)$ and $\chi(x)\chi(x + \frac{1}{2})=\chi_{[-1,-\frac{1}{2}]}(x),\ $so upon using the definition of the integral, we get
$\chi_{[\frac{1}{2},1]}(-1/2)-\chi_{[\frac{1}{2},1]}(-1)=0-0$
More specifically, observe that $d\left(\chi(x)\chi(x + \frac{1}{2})\right)$ refers the measure $\mu$, defined by $\mu ([-1,x))=\chi(x)\chi(x + \frac{1}{2})$, which as we just saw, is equal to $\chi_{[-1,-\frac{1}{2}]}(x)$
Now, $\chi_{[\frac{1}{2},1]}$ is a simple function, so by definition of the Lebesgue-Stieljes integral,
$\int _{[-1,1]}\chi_{[\frac{1}{2},1]}d\mu=1\cdot\mu([\frac{1}{2},1])=1\cdot \mu ([-1,1])-\mu ([-1,\frac{1}{2}])=\chi_{[-1,-\frac{1}{2}]}(1)-\chi_{[-1,-\frac{1}{2}]}(\frac{1}{2})=0-0=0.$
Remark: this all makes intuitive sense since $\mu$ is concentration in the zero set of $\chi_{[\frac{1}{2},1]}(x)$