Doubt about Probability measure in Lebesgue Inegral exercise. I am working my way through the book Basic Stochastic Processes from Springer Undergraduate Mathematics Series and there is a step in one of the excerices' solutions that I can not understand.
The exercice 1.9 goes like this:
Show that if $\eta: \Omega \rightarrow [0,\infty)$ is a non-negative square integrable random variable, then
$E(\eta²)=2\int_0^\infty tP(\eta>t)dt.$
The solution starts as follows:
Let $F(t)=P\{\eta\leq t\}$ be the distribution function of $\eta$. Then
$E(\eta²)=\int_0^\infty t²dF(t).$
Since $P(\eta>t)=1-F(t)$, we need to show that
$\int_0^\infty t² dF(t) = 2\int_0^\infty t(1-F(t))dt$
The next step and the one I do not understand is this one:
$\int_0^a t² dF(t) = \int_0^a t² d(F(t)-1)$
After this it goes on with integration by parts which I can follow.
Any help would be appreciated, I do not understand how $dF(t)$ seems equivalent to the same probability measure minus 1. I am obviously missing something.

$dF(a,b]=F(b)-F(a)$.
The incrementa define the integral with respect yo $dF(t)$.
If we call$$ G(t):=F(t)-1 $$then the stieljes integrals with respect yo F and G Will ve the same because $dF(a,b] $clearly is equal to $ dG(a,b].$
A sidenote: the notation $ d(F(t)-1) $ can ve tricky, You are not translating the measures , maybe that was the confusión.
