$\{ x\in[n,n+1) : m(f^{-1}(\{x\}))>0\}$ has measure zero I am trying to show that if $f: \mathbb{R} \to \mathbb{R}$ is measurable, then the set $S=\{x\in \mathbb{R} : m(f^{-1}(\{x\}))>0\}$ has measure zero where $m$ denotes the Lebesgue measure.
We can show that for each integer $n$, $S\bigcap [n,n+1]=\{x\in [n,n+1]: m(f^{-1}(\{x\}))>0\}$ has measure zero. But I got stuck here.
can someone help me?
Thanks
 A: Define
$$A_{k,n} := \left\{x \in \mathbb{R}; m \big(\{y \in [k,k+1); f(y)=x\} \big)> \frac{1}{n} \right\}.$$
Since for any $x \neq z$ the sets $$\{y \in [k,k+1); f(y)=x\} \quad \text{and} \quad \{y \in [k,k+1); f(y)=z\}$$ are disjoint, there can be at most $n$ points $x_1,\ldots,x_n \in \mathbb{R}$ such that $$m(\{y \in [k,k+1); f(y)=x_j\})>\frac{1}{n};$$ thus $\sharp A_{k,n} \leq n$. This implies that
$$S_n := \{x \in \mathbb{R}; m(f^{-1}(\{x\})> \frac{1}{n}\} = \bigcup_{k \in \mathbb{Z}} A_{k,n}$$
is countable, and therefore
$$S = \bigcup_{n \geq 1} S_n$$
is countable; in particular $m(S)=0$.
A: The set $\{x : \mu(f^{-1}(\{x\}))>0\}$ has to be countable (and hence have measure $0$)because otherwise we would get uncountably many disjoint subsets of positive measure, which is impossible in a $\sigma$-finite measure space. Why is it impossible, you ask. Well, if we have such a uncountable collection $\mathcal{A}$, note that for each $A\in\mathcal{A}$, $A=\bigcup\limits_{N=1}^{\infty}A\cup[-N,N]$, so $\lim_{n \rightarrow \infty}\mu(A\cap[-n,n])=\mu(A)>0$ hence we can find a $N_A\in \mathbb{N}$ s.t. $\mu(A\cap[-N_A, N_A])>0$. If we let $\mathcal{C}$ denote the collection of these $A\cap[-N_A, N_A]$ each $C\in \mathcal{C}$ is obviously contained in a set of the form $[-N,N]$ with $N\in \mathbb{N}$. Letting $\mathcal{C_n}:=\{C\in \mathcal{C} | C\subset [-n,n]\}$ we see that $\mathcal{C}=\bigcup\limits_{n=1}^{\infty}\mathcal{C}_n$. It is impossible that all $\mathcal{C}_n$s are countable for then $\mathcal{C}$ being a countabe union of countable sets would also be countable. Let then $C_M$ be a set that is uncountable. The sets in $\mathcal{C}_M$ are all disjoint, have positive measure and are contained in $[-M, M]$. If we set $\mathcal{D}_n=\{D\in\mathcal{D}|\mu(D)\geq 1/n\}$ we see that $\mathcal{D}=\bigcup\limits_{n=1}^{\infty}\mathcal{D}_n$ and so by the same argument as before there must exist a $\mathcal{D}_N$ is uncountable. By construction $\bigcup\limits_{D\in\mathcal{D}_n} D$ must have infinite measure, but this is impossible as $\bigcup\limits_{D\in\mathcal{D}_n} D \subset [-M,M]$.
