Countable subset of range of $\mu$-measurable function I am reading through the proof of the following. 

If $f\colon X\to\overline{\mathbb R}$ is $\mu$-measurable for some $\sigma$-finite measure $\mu$, then the set $\{r\in\mathbb R\mid \mu(\{x\mid f(x)=r\})>0\}$ is at most countable.

Define $R_{i,k}=\{r\in\mathbb R\mid \mu(\{x\mid f(x)=r\}\cap A_i)>1/k\}$ where $A_i$ are measurable with $\mu(A_i)<\infty$ for all $i$ and $\cup_i A_i= X$. This is at most finite. So $\cup_{i,k} R_{i,k}$ is at most countable and equals the set we are supposed to prove that is countable.
I don't understand either of the bold statements: the first one I can't see it at all while the second one I do understand intuitively  but have no idea how to prove it. 
Any ideas on how to prove them?
 A: Let
$$
A_{i,r}=\{x\mid f(x)=r\}\cap A_i.
$$
Since $A_{i,r}\cap A_{i,s}\ne\emptyset$ for $r\ne s$, if the set $R_{i,k}$ was infinite, then
$$
\infty>\mu(A_i)\ge \mu\left(\bigcup_r A_{i,r}\right)\ge\infty\cdot\frac1k=\infty.
$$
This contradiction shows that the set $R_{i,k}$ is finite.
For the second statement, note that the set $\{x\mid f(x)=r\}$ has positive measure if and only if $\mu(\{x\mid f(x)=r\}\cap A_i)>0$ for some $i$, and that a number $x$ is positive if and only if $x>1/k$ for some $k$. This shows that 
$$
\{r\in\mathbb R\mid \mu(\{x\mid f(x)=r\})>0\}=\bigcup_{i,k}R_{i,k}.
$$
Of course, the union is a countable set because each $R_{i,k}$ is finite.
A: First Part.
Fix $i$ and $k$. Suppose $R_{i,k}$ is infinite, and pick an infinite sequence $(r_n)_{n=1}^\infty$ of distinct elements of $R_{i, k}$. That is, for each $n$ we have $\mu(\{x \mid f(x) = r_n\} \cap A_i) > 1/k$. Then we have
$$
\bigcup_{n=1}^\infty \big(\{x \mid f(x) = r_n\} \cap A_i\big)
\subseteq A_i,
$$
where the union is a disjoint union, so by monotonicity and $\sigma$-additivity we have
$$
\begin{aligned}
\mu(A_i)
&\geq \mu\left(\bigcup_{n=1}^\infty \big(\{x \mid f(x) = r_n\} \cap A_i\big)\right) \\
&= \sum_{n=1}^\infty \mu\big(\{x \mid f(x) = r_n\} \cap A_i\big)
> \sum_{n=1}^\infty \frac{1}{k}
= \infty,
\end{aligned}
$$
contradicting $\mu(A_i) < \infty$. Thus, $R_{i, k}$ is finite.
Second Part.
A real number $x$ is positive if and only if there exists a positive integer $k$ such that $x > 1/k$. This implies that for each $i$ we have
$$
\left\{r \in \mathbb{R} \,\,\middle|\,\, \mu\big(\{x \mid f(x) = r\} \cap A_i\big) > 0\right\}
= \bigcup_{k=1}^\infty \left\{r \in \mathbb{R} \,\,\middle|\,\, \mu\big(\{x \mid f(x) = r\} \cap A_i\big) > \frac{1}{k}\right\}.
$$
Next, since $X = \bigcup_{i=1}^\infty A_i$, a measurable subset of $X$ has positive measure if and only if it has positive measure when intersected with some $A_i$. Thus,
$$
\begin{aligned}
\left\{r \in \mathbb{R} \,\,\middle|\,\, \mu\big(\{x \mid f(x) = r\}\big) > 0\right\}
&= \bigcup_{i=1}^\infty \left\{r \in \mathbb{R} \,\,\middle|\,\, \mu\big(\{x \mid f(x) = r\} \cap A_i\big) > 0\right\} \\
&= \bigcup_{i=1}^\infty \bigcup_{k=1}^\infty \left\{r \in \mathbb{R} \,\,\middle|\,\, \mu\big(\{x \mid f(x) = r\} \cap A_i\big) > \frac{1}{k}\right\}.
\end{aligned}
$$
A: Proof We set  $I=\{r:\mu(\{y:f(y)=r\}>0\}$. Assume the contrary and let $card(I)\ge \aleph_1$. Since $\mu$ is $\sigma$-finite there is a measurable partition $\{A_k: k \in N\}$  of $R$ such that $0\le\mu(A_k)<\infty$ for $k \in N$. Hence there is $k_0$ such that $card(A_{k_0} \cap I)\ge \aleph_1$ and  $0<\mu(A_{k_0})<\infty$.  Now let define a function $F(x)=\mu((-\infty,x)\cap A_{k_0})$ for $x \in R$. It is obvious that $F$ is a monotonous function in $R$. Note that each point $z \in   A_{k_0} \cap I$ is a jump point for $F$. But the cardinality of points of jumps of an arbitrary monotonous  function is at most countable and we get the contradiction.
