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 $A \in \alpha$.
  
  
*
  
*(a) Prove that $ \mu $ is a measure on $ \alpha$.
(b) Is $ \mu $ $ \sigma$-finite ? Is $ \mu$ finite?
  
*(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.
Thank's in advance!
Edit: This was a problem in a examination on measure theory.
 A: 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 $2^{\aleph_0}$
A: 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
$\mathfrak c=2^{\aleph_0}$.)
