Set $f:[0,1]\rightarrow \mathbb{R}$ for which each set $X_{c}=\{x\in [0,1] \mid f(x)\leq c \}$ has Lebesgue measure 0 or 1. Prove that there is $c_{0}$ such that $f(x)=c_{0}$ q.t.p $x\in[0,1]$.
I take $c_{0}:=\inf \{c , m(X_{c})=1 \}$, where $m$ is the Lebesgue measure.
Any help would be appreciated. Thanks!
On
Hint Prove $C:=m(f^{-1}((-\infty, c_0)))=0$ and then use $m(f^{-1}(\{c_0\}))=m(X_{c_0}) -C =1$.
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Let $\phi(c) = m X_c$. Then $\phi$ is non decreasing and takes both values $0,1$. Furthermore, $\phi $ is right continuous, and hence $c_0 = \min \{ c | \phi(c) = 1 \}$ is well defined.
In particular, $m X_{c_0} = 1$ and $m X_{c_0 -{1 \over n}} = 0$ for all $n$. Hence $m \{ x | f(x) = c_0 \} = m ( X_{c_0} \setminus \cup_n X_{c_0 -{1 \over n}} ) = 1$.
On
Let $c_0=\inf\{c:m(X_c)=1\}$, as you said.
For all $c>c_0$, $m(X_c)=1$. Let $c_n$ be a sequence decreasing to $c_0$.
Since $\bigcap_{n} X_{c_n}=\{x:f(x)\le c_0\}$ (think about why this is true), it follows that $$m(\{x:f(x)\le c_0\})=\lim_{n\to\infty} m(X_{c_n})=\lim_n 1=1.$$
For all $c<c_0$, $m(X_c)=0$. Let $c_n$ be a sequence increasing to $c_0$.
Since $\bigcup_{n} X_{c_n}=\{x:f(x)<c_0\}$ (think about why this is true), it follows that $$m(\{x:f(x)<c_0\})=\lim_{n\to\infty} m(X_{c_n})=\lim_n 0=0.$$
Conclude by using $$ m(\{x:f(x)=c_0\})=m(\{x:f(x)\le c_0\})-m(\{x:f(x)<c_0\}). $$
For any $c<c_0$ one has $m(X_c)=0$. Let $q_i, i \in \mathbb{N}$, be an increasing sequence of numbers approaching $c_0$. Then $$\{x: f(x)<c_0\}=\bigcup_{i=1}^\infty \{x: f(x) \leq q_i\}=\bigcup_{i=1}^\infty X_{q_i},$$ and since $m(X_{q_i})=0$, one has $m\{x: f(x)<c_0\}=0$.
Next, let $p_i$ be a decreasing sequence of numbers approaching $c_0$, and let $p_1=\infty$. Then $$\{x: f(x)>c_0\}=\bigcup_{i=1}^{\infty} \{x: p_{i+1}<f(x) \leq p_i\}.$$ But $$m\{x: p_{i+1}<f(x) \leq p_i\}=m\{x: f(x) \leq p_i\} - m \{x: f(x) \leq p_{i+1}\}=1-1=0.$$ It follows that $m \{x: f(x)>c_0\}=0$. Therefore, $m\{x: f(x) \neq c_0\}=m\{x: f(x)<c_0\}+m\{x: f(x)<c_0\}=0$ which implies the claim.