Lebesgue measure on a continuous function to prove equality

Question

Let $f(x)$ be a continuous function on [−1, 1].

Show that there exists a constant $c$ such that the Lebesgue measures $\mu (\{x ∈ [−1, 1] : f(x) ≥ c\}) ≥ 1$,$\quad$ $\mu (\{x ∈ [−1, 1] : f(x) ≤ c\}) ≥ 1$.

This part has been done here. Lebesgue measure on a continuous function

Now I want to extend the result to prove the following.

For this constant $c$, prove the equality

$\int_{-1}^{1} |f(x) − c| dx$ = $min_{k\in R}$ $\int_{-1}^{1} |f(x) − k| dx$

Answer 1

Suppose $k>c$. For every $x$ the triangle inequality shows that $$|f(x)-k|\ge|f(x)-c|-|k-c|=|f(x)-c|+c-k.$$On the other hand, if $f(x)\le c$ then $$f(x)-k=(f(x)-c)+(c-k),$$ and since $f(x)-c\le0$ and $c-k\le0$ this shows $$|f(x)-k|=|f(x)-c|+k-c\quad(f(x)\le c).$$Let $A=\{f(x)\le c\}$ and $B=[-1,1]\setminus A$. So $m(A)\ge1$ and $m(B)\le 1$. Now $$\int_{-1}^1|f-k| \ge\int_A\left(|f-c|+(k-c)\right)+\int_B(|f-c|+(c-k))=\int_{-1}^1|f-c|+(m(A)-m(B))(k-c).$$Similarly for $k<c$.