If a function is integrable, does it also have finite integral given any counting measure? If the integral of a function $f$ is $L^1$,
\begin{equation} \int f dx < \infty \end{equation}
Does the same function have finite integral under any countable counting measure,
\begin{equation} \sum f(x_{i}) d\mu(x_{i}) < \infty \end{equation}
for any sequance $\{ x_{i} \}_{i \in \mathbb{N}}$?
I wanna say yes, due to the latter being the same as simple functions on point sets. But also we might loose cancelation of areas when throwing away alot of stuff in this fashion.
 A: So, first of all, please make sure you get your notation correct, otherwise it's hard to be sure what you're asking.  I think you mean to say that $f\in L_1$, not $L_2$.  It would also be nice to have the domain of $f$, which I assume from context is $(0,\infty)$.  Also, if you want to talk about an integral over the counting measure on $\mathbb{N}$, the proper notation is
$$\sum_{i=1}^\infty f(x_i).$$
Of course, the answer is no.  Consider the function $f=\boldsymbol{1}_\mathbb{N}$.
There are special conditions under which the answer is yes.  By the Integral Test, if $f$ is nonnegative, continuous, and decreasing on $(0,\infty)$ and $\int f<\infty$ then $\sum_{n=1}^\infty f(n)$ converges.  In fact, so does $\sum_{i=1}^\infty f(x_i)$, provided $(x_i)_{i=1}^\infty$ is increasing with $\inf|x_{i+1}-x_i|>0$.
A: If $f:\mathbb R\to\mathbb R$ is defined as $f=\mathbf1_{\mathbb Z}$ then $\int f(x)^2dx=0$ but $\sum_{n\in\mathbb Z}f(n)=\infty$.
A: Your question is confusing, but I think the answer is essentially, "no".
You can change the value of an integrable function on any countable set of points without changing integrability or the integral. So just define it to be large enough at each point at which your "counting measure" is supported to make the sum diverge.
