# Number of elements of $A_k = \{ B \subset \mathbb N : \mu(B) = k \}$, $k \in \mathbb R$

Let $X= \mathbb N$ , $\alpha = \mathcal{P} (\mathbb N)$, and $u_n =\frac{1}{n+1}$ for every $n \in \mathbb N$.

Define the function $\mu$ on $\alpha$ by $\mu(A)=\sum\limits_{n \in A} u_n$ for every $A \in \alpha$.

1. (a) Prove that $\mu$ is a measure on $\alpha$.

(b) Is $\mu$ $\sigma$-finite ? Is $\mu$ finite?

2. (a) Find the image of $\mu$.

(b) Is $\mu$ one-to-one?

(c) If $A_k = \{ B \subset \mathbb N : \mu(B) = k \}$ with $k \in \mathbb R ^{+}$, find the number of elements of $A_k$.

I need help only for 2 (c). I have done the others.

Edit: This was a problem in a examination on measure theory.

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I think your definition of $u_n$ is wrong. You probably want $u_n = \frac{1}{n+1}$. –  Quinn Culver May 17 '12 at 18:50
@QuinnCulver: But that I have define as $u_n$. –  passenger May 17 '12 at 18:56
I think that every $A_k$ for $k>0$ is infinite. Is this what you want proof, or do you need the exact cardinality of $A_k$? –  Martin Sleziak May 17 '12 at 19:05
@MartinSleziak: Can you explain first why every $A_k$ is infinite? –  passenger May 17 '12 at 19:07
I hope the original question was not a verbatim of your examination text. –  Did May 17 '12 at 20:19

For any $k>0$, $A_k$ has cardinality $\mathfrak{c} = 2^{\aleph_0}$ : let $N_0$ be the set of even integers and $N_1$ be the set of odd integers. Since $\sum_{n \in N_0} u_n = \sum_{n \in N_1} u_n = \infty$, it is again the case that you can obtain any real as a measure of a subset of $N_0$ or as a subset of $N_1$ : $\mu( \mathcal P (N_0)) = \mu( \mathcal P (N_1)) = [0, + \infty]$.

Let $0<l<k$, and choose a subset $X_0$ of $N_0$ such that $\mu(X_0) = l$, and a subset $X_1$ of $N_1$ such that $\mu(X_1) = k-l$. Then, $\mu(X_0 \cup X_1) = l+(k-l) = k$. Since for any fixed value of $k$ we can choose $2^{\aleph_0}$ many values for $l$, we obtain $2^{\aleph_0}$ many distincts subsets $X$ of $\mathbb N$ such that $\mu(X)=k$.

Meanwhile, $A_k$ is at most of cardinality $2^{\aleph_0}$ since it is a subset of $\mathcal P(\mathbb N)$. Thus $A_k$ has cardinality $2^{\aleph_0}$

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Nice clear solution mercio! Thank you very much! –  passenger May 17 '12 at 19:47

I suppose that in the preceding part of the exercise you have shown that image of $\mu$ is $[0,\infty)$, i.e. every non-negative real number can be be obtained as $\mu(A)$ for some set $A$.

More generally, for arbitrary decreasing sequence $u_n$ we have, that every positive number can be obtained as $\sum_{n\in A}u_n$ whenever $\sum_{k=1}^\infty u_k=\infty$ and $\lim\limits_{k\to\infty} u_k=0$, see e.g. here. (The claim given in that answer is more general than this.) You have shown this for $u_n=\frac1{n+1}$ in the preceding part of this exercise, it should be easy to check that the same argument works for arbitrary such sequence.

It is obvious that $A_0=\{\emptyset\}$. Now we can show that if $x>0$ can be obtained as $\mu(A)=x$, then there is infinitely many such sets, i.e. each $A_x$ is infinite.

Indeed, let $x$ has a representation $\sum_{n\in A} u_n$ and $n_0\in A$. Then we can simply omit $u_{n_0}$ from the sequence and apply the same observation to the new sequence. Again, $x$ has some representation, but it has to be a different one, since it does not contain $n_0$.

We can continue like this - we choose some $n_1$, then we obtain a new representation which does not contains $n_0$, $n_1$ etc. We will get infinitely many sets belonging to $A_x$.

However, I don't see what cardinality will the set $A_x$ have. (Aside from obvious estimate that it is between $\aleph_0$ and $\mathfrak c=2^{\aleph_0}$.)

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Here: "Now we can show that if x>0 can be obtained as μ(A)=x, then there is infinitely many such sets, i.e. each Ax is infinite." what did you take as A ? –  passenger May 17 '12 at 19:26
I took arbitrary $A$ with this property. At least one $A$ must exist for each $x$ in the range of $\mu$. (You described the range in the part 2(a).) –  Martin Sleziak May 17 '12 at 19:29
Why $A_k$ is infinte? I can't still see it. –  passenger May 17 '12 at 19:33
The basic observation is that for arbitrary $N$ you can find a set $A\in A_x$ such that $A\subseteq\{N,N+1,N+2,\ldots\}$. In this way, you can alway get new sets. (This is slightly different from what I wrote in the answer, but the basic idea is the same.) –  Martin Sleziak May 17 '12 at 19:37
What do you mean here: "$A\subseteq\{N,N+1,N+2,\ldots\}$" ? –  passenger May 17 '12 at 19:42