Additivity and $\sigma$-Additivity of a Measure on $\mathbb{N}$ with Example

[LHeiner]

Hey

i'm given a measure on $\Omega =\mathbb{N}, A \subset \mathbb{N}$ and should check additivity as well as $\sigma$-additiviy. Additionally i should give an example for a set where $\mu(A)$ is not defined:

$\mu(A)=\lim_{n \rightarrow \infty} \frac{1}{n} \xi\mid _\mathbb{N} (A \cap [1,n])$ with finite limit.
whereas $\xi\mid _\mathbb{N}$ is the counting measure on $\mathbb{N}$

Can anybody help me with this?
Appreciate it, thanks very much

 

[micromass]

For the additivity

\mu(A\cup B) =\lim_{n\rightarrow +\infty}{\frac{1}{n}\xi_\mathbb{N}((A\cap [1,n])\cup(B\cap [1,n]))}

Now just apply the additivity of the counting measure and split the limit.

For \sigma-additivity, you will eventually need to exchange limit and series. But this should be no problem since the series has positive terms.

As for the example where \mu does not exist. The only problem with \mu might be that the limit does not exist. This happens with the following set
\{n\in \mathbb{N}~\vert~n>0\}\cup \{2n+1/2~\vert~n>0\}

Just ask if something is not clear...

The measure \mu can actually be regarded as the limiting average elements of a set. It also poses some interesting questions. Like: what is the measure of the primes? or what is the measure of the squares?

 

[bpet]

A nice example is discussed here: https://www.physicsforums.com/showthread.php?t=441898.

 

[LHeiner]

Ok that means:

$\mu (A\cup B)=\lim_{n\rightarrow \infty} \frac{1}{n} \xi\mid_\mathbb{N}(A\cap[1,n])\cup (B\cap[1,n]))=\\<br /> \lim_{n\rightarrow \infty} \frac{1}{n} \xi\mid_\mathbb{N} [(A\cap[1,n])+ (B\cap[1,n])]=\\<br /> \lim_{n\rightarrow \infty} \frac{1}{n} \xi\mid_\mathbb{N} (A\cap[1,n])+\lim_{n\rightarrow \infty} \frac{1}{n} \xi\mid_\mathbb{N} (B\cap[1,n])=\mu(A)+\mu(B)$

and $\sigma$-add:

$\mu(\bigcup_{i=1}^{n}A_i)=\lim_{n\rightarrow \infty} \frac{1}{n} \xi\mid_\mathbb{N} (\bigcup_{i=1}^{n}A_i \cap [1,n])\\<br /> =\lim_{n\rightarrow \infty} \frac{1}{n} \xi\mid_\mathbb{N} (\sum_{i=1}^{n}A_i \cap [1,n])\\<br /> =\sum_{i=1}^{n} \lim_{n\rightarrow \infty} \frac{1}{n} \xi\mid_\mathbb{N} (A_i \cap [1,n])= \sum \mu(A_i)$

But i can't see why $\mu$ has no limit for the set ${n \in \mathbb{N}\vert n&gt;0}\cup \{2n+1/2\vert n&gt;0}$

 

[bpet]

The "measure" isn't sigma-additive, e.g. consider a countable union of singletons.

 

[micromass]

Hmm, in your proof of additivity, you say

\xi[(A\cap [1,n])+(B\cap[1,n])]

You can of course not take the sum of two sets. That needs to be

\xi(A\cap [1,n])+\xi(B\cap [1,n])

As noticed by bpet, \mu is not \sigma-additive. The reason for this is, that you can not in general exchange limit with sum. This is sometimes allowed, but not in this case.

For convenience, I'll give two situations in which you CAN exchange limit with sum (notice that none of these two situations apply here).

We have that

\lim_{n\rightarrow +\infty}{\sum_{k=1}^{+\infty}{x_{n,k}}=\sum_{k=1}^{+\infty}{\lim_{n\rightarrow +\infty}{x_{n,k}}}

if one of following situations occurs
1) \forall k: m\leq n~\Rightarrow~x_{m,k}\leq x_{n,k} and every x_{n,k} is positive.
2) There exists a sequence (y_k)_k such that \sum_{k=1}^{+\infty}{|y_k|}&lt;+\infty and \forall k,n: |x_{n,k}|\leq y_k

You will prove these theorems later in it's ful generality (i.e. with integrals instead of sums).




Now for the set A=\{n \in \mathbb{N}\vert n&gt;0\}\cup \{2n+(1/2)\vert n&gt;0\}. Just calculate for every n, how many elements are in A\cap[1,n]. (be sure to discriminate between the cases n is even/n is odd). Then divide this number by n and take the limit.

 

[LHeiner]

I'm sorry now I am completely confused

i think my problem is that maybe i don't completely understand the function at all. i mean isn't $\mu(A)$ always 0?

If i look at A\left \{1,2,5,8,9,10 \right \} \Rightarrow \mu(A)=\lim_{n\rightarrow \infty} \frac{1}{n} \xi (A \cap [1,n])= \lim_{n\rightarrow \infty} \frac{1}{n} \xi ({1,2,5,8,9,10})=\lim_{n\rightarrow \infty} \frac{6}{n}=0

i think measure (&set) theory is not my speciality

 

[micromass]

Don't worry, everybody is confused in the beginning of measure theory. It grows on you.

But the measure \mu is certainly not zero. For finite sets, it is zero. But let's calculate \mu(\mathbb{N}).

We have that \mathbb{N}\cap [1,n]=\{1,...,n\}. So \xi(\mathbb{N}\cap [1,n])=n. So

\frac{1}{n} \xi(\mathbb{N}\cap [1,n])= 1

Taking the limit, yields that \mu(\mathbb{N})=1.

 

[LHeiner]

okey so i think i now get why the function is not \sigma additive, because if you apply the function to sum of countable unions of singletons you don't get the same as if you sum up the function on each of the sets.
am i right?

so then i only have to figure out the last part the set for which there exists no limit

 

[micromass]

Yes, you are right.

 

[LHeiner]

still confused with this set, why is there no limit, even if i distinguish between odd/even i always get a countable set or where is my error?

n..even= 2k
A=\left \{2k\right \}\cup \left \{4k+0,5\right \}=\left \{2,4,4.5,6,8,8.5...\right \}\cap [1,n]\rightarrow \xi(..)=n
n..odd=2k+1
A=\left \{2k+1\right \}\cup \left \{4k+1,5\right \}=\left \{1,3,5,5.5,7...\right \}\cap [1,n]\rightarrow \xi(..)=n

 

[micromass]

OK, maybe you should try bpet's example https://www.physicsforums.com/showthread.php?t=441898. It is nicely worked out there.

I'm also doubting if my example would really work...