Lebesgue measure on a continuous function Let $f(x)$ be a continuous function on [−1, 1].
Show that there exists a constant $c$ such that the Lebesgue measures
$\mu (\{x ∈ [−1, 1] : f(x) ≥ c\}) ≥ 1$,$\quad$ $\mu (\{x ∈ [−1, 1] : f(x) ≤ c\}) ≥ 1$.
I can do the trivial case choosing $c$ to be the supremum and infimum of $f$ over the interval. But I think this is not a kosher proof.
 A: Since $f$ is continuous, let $m$ and $M$ be the minimum and maximum of $f$ on $[-1,1]$. Consider two functions $g,h:[m,M]\rightarrow\mathbb{R}$ such that 
\begin{align*}
  g(c)&=\mu(f\geq c),\\
  h(c)&=\mu(f\leq c).
\end{align*}
Clearly, $g$ is nonincreasing and $h$ is nondecreasing. Moreover, we have $g(m)=h(M)=2$. If either $g(M)\geq 1$ or $h(m)\geq 1$ holds, we are done. Now assume $g(M)<1$ and $h(m)<1$. 
If $c_n\uparrow c$, then $f(x)\geq c_n$ for all $n$ if and only if $f(x)\geq c$. Hence $\cap_n(f\geq c_n)=(f\geq c)$. This implies $g$ is left continuous. Similarly, $h$ is right continuous. Define $\tilde{c}=\sup\{c|g(c)\geq 1\}$. Because $g$ is nonincreasing and left continuous, $g(\tilde{c})\geq 1$. Define $\hat{c}=\inf\{c|h(c)\geq 1\}$. Because $h$ is nondecreasing and right continuous, $h(\hat{c})\geq 1$.
If $\hat{c}>\tilde{c}$, then for any $c\in(\tilde{c},\hat{c})$, $g(c)<1$ and $h(c)<1$. But this is impossible since $g(c)+h(c)\geq 2$ for all $c$. Hence $\hat{c}\leq\tilde{c}$. Then for any $c\in[\hat{c},\tilde{c}]$, $g(c)\geq 1$ and $h(c)\geq 1$.
