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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 $\{5\}$?

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  • $\begingroup$ The geometric series is $2$ if you start at $n=0$ otherwise $1$. So does your $\Bbb N$ include $0$ or not? $\endgroup$ – Henno Brandsma Sep 21 at 8:07
  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Sep 21 at 8:11
  • $\begingroup$ @HennoBrandsma You are right. I should start in 1 and therefore remove n=0 $\endgroup$ – Daniel Sep 21 at 8:14
  • $\begingroup$ Easier: sum of geometric series equals first term divided by (one minus ratio): so $\frac12$ divided by $\frac12$ so $1$. $\endgroup$ – Henno Brandsma Sep 21 at 8:18
  • $\begingroup$ @HennoBrandsma Very nice explanation regarding the Dirac measure, thank you $\endgroup$ – Daniel Sep 21 at 8:27
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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.

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