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.

Solution 1:

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.