Ytukyg

I have a similar question as here.

Lesbesgue-Stieltjes measure $mu_F$ with the function $F=lvert xrvertlfloor xrfloor$. i.e. $mu_F([a,b))=F(b)-F(a)$. I am trying to find the Lebesgue-Radon-Nikodym decomposition of $mu_F$ with respect to the Lebesgue measure $m$.

Now, I know this function has countably many discontinuities at each integer-valued $x$, so I believe I want the mutually singular part to be the Dirac Measure of each integer, and take the absolute continuous part to be the Lebesgue measure itself. But it would have to be a countable sum of Dirac measures for the discontinuity points… and I am not sure if I am allowed to do that, or if my approach is ok.

I would love to get some feedback/a point in the right direction or any help in general.

$defd{mathrm{d}}$It is fine to add countably many measures to define a new measure as long as all of them are non-negative.

For this question, define$$
μ_c(d x) = |lfloor xrfloor| , m(d x),quad μ_s(d x) = sum_{n in mathbb{Z}} |n| δ_n(d x),
$$
i.e.$$
μ_c(A) = int_A |lfloor xrfloor| ,d x,quad μ_s(A) = sum_{n in mathbb{Z}} |n| I_A(n).
$$
It easy to verify that $μ_c$ and $μ_s$ are indeed measures and $μ_c ll m$, $μ_s ⊥ m$.

To prove that $μ_F = μ_c + μ_s$, it suffices to prove that for any $k in mathbb{Z}$ and $k leqslant a < b leqslant k + 1$,$$
μ_c((a, b]) + μ_s((a, b]) = μ_F((a, b]).
$$
If $b < k + 1$, note that $ab > 0$, then$$
μ_c((a, b]) + μ_s((a, b]) = |k| (b - a) + 0 = |b| k - |a| k = μ_F((a, b]).
$$
Otherwise, note that $|k| (k + 1) = |k + 1| k$ since $k in mathbb{Z}$, thenbegin{align*}
&mathrel{phantom{=}}{} μ_c((a, k + 1]) + μ_s((a, k + 1]) = |k| (k + 1 - a) + |k + 1|\
&= |k + 1| (k + 1) - |k| a = μ_F((a, b]).
end{align*}
Therefore, $μ_F = μ_c + μ_s$.