Integration with respect to the Lebesgue-Stieltjes measure associated with floor function For two different measures, I am trying to identify $L^1(\mu)$(the set of integrable functions, i.e. $\{f:\int \lvert f\rvert du<\infty\}$, and trying to compute $\int fd\mu$ for $f\in L^1(\mu)$.


*

*on $\mathbb R, \mu=\mu_F$, the Lebesgue-Stieltjes measure with $F(x)=\lfloor x\rfloor$.


Attempts 


*

*I know that the unique measure derived from the floor function would measure the number of integers between the end points, or in the set. But I'm not sure how to build on that to use to calculate the integral for a general $f$, or what would make a function integrable.


Update: $$\int 1_{E}(x)d\mu(x)=\mu(E)=\sum_{n\in\mathbb Z}1_E(n)$$
should be true for any characteristic function. And for this integral to be finite, in any Borel set $E$, there should only be finite number of integers in the set. But I'm not sure how to bootsrap up from this point to simple functions, and then functions in general.
Any help would be greatly appreciated!
 A: In your update you have correctly identified that $\mu=\sum_{k\in\mathbb{Z}}\delta_k$.
Take a simple function $f(x)=\sum_{i=1}^n a_i \chi_{A_i}(x)$ such that $\mu(A_i)<\infty$ for all $i$. Its integral is defined as
$$
\int f\,d\mu = \sum_{i=1}^n a_i \mu(A_i)
= \sum_{i=1}^n a_i \sum_{k\in\mathbb{Z}} \delta_k(A_i)
= \sum_{k\in\mathbb{Z}} \sum_{i=1}^n a_i \delta_k(A_i)
= \sum_{k\in\mathbb{Z}} \sum_{i=1}^n a_i \chi(A_i)(k)
= \sum_{k\in\mathbb{Z}} f(k) .
$$
If now $f$ is a non-negative Borel function and $(f_n)$ are simple functions converging monotonically to $f$, you have
$$
\int f\,d\mu = \lim_{n\to\infty} \sum_{k\in\mathbb{Z}} f_n(k)
= \sum_{k\in\mathbb{Z}} f(k)
$$
by monotone convergence.
Such a non-negative function is integrable iff $\sum_{k\in\mathbb{Z}} f(k) < \infty$.
Finally, if $f=f^+-f^-$ is a generic Borel function, it belongs to $L^1(\mu)$ if and only if $\int f^\pm\,d\mu < \infty$ and in such case you have
$$
\int f\,d\mu = \sum_{k\in\mathbb{Z}} f^+(k) - \sum_{k\in\mathbb{Z}} f^-(k)
= \sum_{k\in\mathbb{Z}} f(k) .
$$
So $f\in L^1(\mu)$ iff
$$
\int |f|\,d\mu = \sum_{k\in\mathbb{Z}} |f(k)| < \infty .
$$
