Let $f:A \to \Bbb R^n$ be measurable. Show that $\{x \in \Bbb R^n \mid m(f^{-1}\{x\}) > 0\}$ is a countable set. 
Let $f:A \to \Bbb R^n$ be measurable. Show that $$\{x \in \Bbb R^n \mid m(f^{-1}\{x\}) > 0\}$$ is a countable set.

How can one approach this? I only know few techniques to show that a set is countable. One of them is to show that there is a injection from the set to $\Bbb N$ or that it's measure is zero. Neither of these seem applicable here. It seems that I'm to show that the set of points in $\Bbb R^n$ whose fiber has positive measure is countable set, why should this be true to begin with? $\Bbb R^n$ is an uncountable set so there should be uncountable many points, but apparrently only countable many whose fiber has positive measure?
 A: Note that this only is guaranteed to hold when $A$ has finite measure. WLOG, assume that $A$ has measure 1.
Write $S_i = \{x \mid m(f^{-1}(\{x\})) > 1 / i\}$ for $i \in \mathbb{N}_+$. Then we seek to show $\bigcup\limits_{i \in \mathbb{N}_+} S_i$ is countable, so it suffices to show that each $S_i$ is countable. In fact, each $S_i$ is finite.
Since $m(A) = 1$, I claim that $|S_i| < i$. For suppose we could find distinct elements $x_1, \ldots, x_i\in S_i$. Then we would have $1 < m(f^{-1}(\{x_1\})) + \ldots + m(f^{-1}(\{x_i\})) = m(\bigcup\limits_{j = 1}^i f^{-1}(x_i)) \leq m(A) = 1$, which is a contradiction.
Why do we need to assume $A$ has finite measure? Because the theorem turns out to be false otherwise. Consider $\mathbb{R}$ with the cardinality measure
$$m(S) = \begin{cases}
  |S| & S \text{ is finite} \\
  \infty & otherwise
\end{cases}$$
defined on the $\sigma$-algebra $P(\mathbb{R})$.
Then $f : \mathbb{R} \to \mathbb{R}$ is measurable, and $\{x \mid m(f^{-1}(\{x\})) > 0\} = \mathbb{R}$ is uncountable.
A final note: for most practical cases, we can get around the requirement that $A$ have finite measure. For example, consider $\mathbb{R}^n$, and take some continuous function $f : \mathbb{R}^n \to \mathbb{R}$ which is positive everywhere and has finite integral (for instance, $f(x_1, \ldots, x_n) = e^{-(x_1^2 + \cdots + x_n^2)}$). Then we can define a measure $m'(A) = \int \chi_A f d\mu$, where $\mu$ is the Lebesgue measure. Note that for all $A$, $m'(A) > 0$ iff $\mu(A) > 0$. So the theorem also holds for any $A \subseteq \mathbb{R}^n$ with the inherited Legesgue measure, since we can use the $m'$ measure instead of the Legesgue measure.
A: Actually, $m$ needs not be the Lebesgue measure and we can work on
the following abstract setting.
Let $(X,\mathcal{F},m)$ be a measure space, where $m$ is a $\sigma$-finite
measure, in the sense that there exists a sequence $(X_{k})$ in $\mathcal{F}$
such that $X=\cup_{k}X_{k}$ and $m(X_{k})<\infty$. Let $f:X\rightarrow\mathbb{R}^{n}$
be a measurable function. Then $\{y\in\mathbb{R}^{n}\mid m\left(f^{-1}\{y\}\right)>0\}$
is a countable set.
Proof: For each $k\in\mathbb{N}$, define $B_{k}=\{y\in\mathbb{R}^{n}\mid m\left(f^{-1}\{y\}\cap X_{k}\right)>0\}$.
Define $B=\{y\in\mathbb{R}^{n}\mid m\left(f^{-1}\{y\}\right)>0\}$.
Observe that $B\subseteq\cup_{k}B_{k}$. Prove by contradiction. Suppose
that there exists $y\in B\setminus\cup_{k}B_{k}$, then $m\left(f^{-1}\{y\}\cap X_{k}\right)=0$
for each $k$. We have that
\begin{eqnarray*}
 &  & m\left(f^{-1}\{y\}\right)\\
 & = & m\left(\cup_{k}\left[f^{-1}\{y\}\cap X_{k}\right]\right)\\
 & \leq & \sum_{k=1}^{\infty}m\left(f^{-1}\{y\}\cap X_{k}\right)\\
 & = & 0,
\end{eqnarray*}
which is a contradiction.
Next, we show that $B_{k}$ is countable for each $k$. Prove by contradiction.
Suppose that $B_{k}$ is uncountable for some $k$. For each $l\in\mathbb{N}$,
define $B_{k,l}=\{y\in\mathbb{R}^{n}\mid m\left(f^{-1}\{y\}\cap X_{k}\right)>\frac{1}{l}\}.$
Since $B_{k}=\cup_{l=1}^{\infty}B_{k,l}$, there exists $l$ such
that $B_{k,l}$ is uncountable. Therefore, it is possible to choose
$y_{1},y_{2},\ldots\in B_{k,l}$ such that $y_{1},y_{2},\ldots$ are
pairwisely distinct. For each $N\in\mathbb{N}$, define $C_{N}=\{y_{1},y_{2},\ldots,y_{N}\}$.
Observe that $f^{-1}\{y_{i}\}\cap f^{-1}\{y_{j}\}=\emptyset$ whenever
$i\neq j$
We have that
\begin{eqnarray*}
m(X_{k}) & \geq & m\left(f^{-1}(C_{N})\cap X_{k}\right)\\
 & = & m\left(\cup_{i=1}^{N}\left[f^{-1}\{y_{i}\}\cap X_{k}\right]\right)\\
 & = & \sum_{i=1}^{N}m\left(f^{-1}\{y_{i}\}\cap X_{k}\right)\\
 & > & \frac{N}{l}
\end{eqnarray*}
which is impossible because $m(X_{k})<\infty$ but $\frac{N}{l}$
can be made arbitrarily large.
