# Integral of counting measure

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 $$\{5\}$$?

• The geometric series is $2$ if you start at $n=0$ otherwise $1$. So does your $\Bbb N$ include $0$ or not? – Henno Brandsma Sep 21 at 8:07
• Did you cover $\int fd(\mu+\delta_2+\delta_5) = \int f d\mu + \int f d\delta_2 + \int f d\delta_5$? That would justify your approach, as $\int f \delta_x = f(x)$ generally. – Henno Brandsma Sep 21 at 8:11
• @HennoBrandsma You are right. I should start in 1 and therefore remove n=0 – Daniel Sep 21 at 8:14
• Easier: sum of geometric series equals first term divided by (one minus ratio): so $\frac12$ divided by $\frac12$ so $1$. – Henno Brandsma Sep 21 at 8:18
• @HennoBrandsma Very nice explanation regarding the Dirac measure, thank you – Daniel Sep 21 at 8:27

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.