I am looking at a homework problem: Measure space ($\mathbb{N}, \mathcal{P}(\mathbb{N}),\mu)$) where $\mu$ is the counting measure.
Let $\nu=\mu+\delta_2+\delta_5$ where $\delta$ is the Dirac measure.
Determine $\int_\mathbb{N} f d\nu$ where $f(n)=2^{-n}$
My solution is to sum over the natural numbers and view it as a geometric series:
$\int_\mathbb{N} f d\nu=(1/2)^1 + (1/2)^2 +...(1/2)^n +1/2^2+1/2^5= 1/(1-1/2)-(1/2)^0+ (1/2)^2 + (1/2)^5=2-1 + (1/2)^2 + (1/2)^5 = 1+1/4+1/32$
But I don't know if it is wrong with the 2 Dirac measures. How do I handle that the measure $\nu$ returns $2$ instead of $1$ on singletons $\{2\}$ and $\{5\}$?
Your approach is valid: for positive measures we can use
$$\int f d (\mu_1 + \mu_2) = \int f d\mu_1 + \int f d\mu_2$$
(all finite sums) and you should know, as basic examples that
$$\inf f d\mu = \sum_{n=1}^\infty f(n)$$
when $\mu$ is the counting measure on $\Bbb N = \{1,2,3,\ldots\}$ and positive functions on it.
and $$\int f d \delta_x = f(x)$$
for Dirac measures $\delta_x$ generally.
Applying those facts plus $\sum_{n=1}^\infty ar^n = \frac{ar}{1-r}$ (for $r<1$) we get your answer.