Does a $\sigma$-finite measure always admit a countable partition whose components are uniformly bounded from below and/or above? Let $(A,\mathcal{A})$ be a measure space and $\mu$ be a $\sigma$-finite
  measure on $A$ such that $\mu(A)=\infty$. 
Is it true that then one can find a
  partition $(A_m)_{m \geq 1}$ of $A$ such that not only $\mu(A_m) <\infty$, but
  also $\mu(A_m) > \delta$ for all $m$, where $\delta$ is some arbitrary
  positive number? What about an upper bound, i.e. can we always find a partition such that $\mu(A_m) < M$ for all $m$ and some arbitrary positive number $M$?
I know that this is a rather elementary question but would still be thankful for a quick assurance that my proof below for the lower bound is correct and some comments for the upper bound.
 A: Here's a proof for the lower bound: First, let $(B_n)_{n \geq 1}$ be a partition of $A$ such that $\mu(B_n) <
  \infty$ for all $n \geq 1$. Then, for $m \geq 0$, inductively define positive
  integers $k_m$ by setting $k_0=0$ 
  and, for $m \geq 1$,
  $$ k_{m+1} = \operatorname{min} \left\{ k > k_m \, \colon \, \mu \left(
      \bigcup_{n=k_{m}+1}^{k} B_{n} \right) > \delta \right\},$$
  where we set $\min \emptyset = \infty$. 
  Assume that the set $\{m \colon k_m = \infty\}$ is not empty and choose its smallest element $m_0$. Then $\mu \left( \bigcup_{n=k_{m_0-1}+1}^{\infty}
    B_n \right) < \delta$ and thus $\mu(A) = \mu \left( \bigcup_{n=1}^{k_{m_0-1}} B_n
  \right) + \mu \left( \bigcup_{n=k_{m_0-1}+1}^{\infty} B_n \right) < \infty$, which
  is a contradiction. Thus $k_m < \infty$ for all $m \geq 0$ and the
  partition $(A_m)_{m \geq 1}$ defined by
  $$A_m = \bigcup_{n=k_{m-1}+1}^{k_m} B_n$$
  has the required property.
A: Let $A$ a set of all natural numbers and $\cal{A}$ be powerset of $A$. Let $\lambda(\{n\})=n$ for each $n \in N$. We define a measure $\mu$ on $\cal{A}$ as follows: $\mu(X)=\sum_{n \in  X}\lambda(\{n\})$. Then $\mu$ is a σ-finite measure on A such that $\mu(A)=+\infty$, there is a partition $(A_m)_{m≥1}$ of $A$ such that not only $\mu(A_m)<\infty$ , but also $\mu(A_m)>\delta$ for all $m$, where $\delta$ is some arbitrary positive number and there is no an upper bound, i.e.  for an arbitrary partition $(B_m)_{m \in N}$ there does not exist a positive number $M$ such that $\mu(B_m)<M$ for all $m$. 
