Let $a_f=\text{ arg} \min_{a} \int \left|f(x)-a\right| dx$ and $a_g= \text{ arg} \min_{a} \int \left|g(x)-a\right| dx$, is $a_f \le a_g$? Let $ f(x) \le g(x) $ and assume that $g(x),f(x) \in L^1$ let 
\begin{align}
a_f= \text{ arg} \min_{a } \int_A \left|f(x)-a\right| dx\\
a_g=\text{ arg} \min_{a } \int_A \left|g(x)-a\right| dx
\end{align}
where $A \subseteq \mathbb{R}$. 
We assume that such $a_f$ and $a_g$ exist and $|a_f|,|a_g| < \infty$.
Is $a_f \le a_g$? How would one approach this problem?
Thank you.
 A: Let $\lambda$ denote the Lebesgue measure.
If $\lambda(A) = \infty$, then both $a_f$ and $a_g$ are zero. Indeed, for $a>0$, 
$$
\lambda(\{x\in A: |f(x)|> a/2\})\le 2 a^{-1}\int_{A} |f(x)|\, dx<\infty,
$$
therefore, 
$$
\int_{A} |f(x)-a|\, dx\ge \frac{a}2\lambda(\{x\in A: |f(x)|\le a/2\}) = \infty. 
$$
(Indeed, $|z-a|\ge \frac{a}2\mathbf{1}_{|z|\le a/2}$.) Similarly, for $a<0$,
$$
\int_{A} |f(x)-a|\, dx = \infty.
$$
Therefore, the infimum is attained for $a=0$: the expression is infinite for other values.

When $0<\lambda(A)<\infty$, any minimizer $a_f$ is the "median" of $f$ in the following sense: $$\lambda(\{x\in A: f(x) < a_f\}) \le \lambda(\{x\in A: f(x) \ge  a_f\})\tag{1}$$ and $$\lambda(\{x\in A: f(x) > a_f\}) \le \lambda(\{x\in A: f(x) \le  a_f\}).\tag{2}$$ 
Indeed, if some of these inequalities fails (say, the first one), then for some $\epsilon >0$ 
$$
\lambda(\{x\in A: f(x) < a_f-\epsilon\}) > \lambda(\{x\in A: f(x) \ge  a_f-\epsilon\}) 
$$
in view of the continuity of probability. Then 
$$
\int_A |f(x) - a_f + \epsilon| dx = \int_{x\in A: f(x) < a_f-\epsilon} (a_f - \epsilon -f(x))dx  + \int_{x\in A: f(x) \ge a_f-\epsilon} (f(x) - a_f + \epsilon)dx\\ 
\le \int_{x\in A: f(x) < a_f} (a_f  -f(x))dx - \epsilon \lambda(\{x\in A: f(x) < a_f-\epsilon\})\\ + \int_{x\in A: f(x)\ge a_f} (f(x) - a_f)dx + \epsilon \lambda(\{x\in A: f(x) \ge  a_f-\epsilon\})\\
< \int_A |f(x) -a_f| dx,
$$
contradicting the minimality.
It is not hard to see (using the continuity of probability) that such $a_f\in [a_{*,f},a^*_f]$, where $a^*_f$ is the supremum of values for which (1) holds, and $a_{*,f}$ is the infimum of values for which (2) holds. It is possible that $a_{*,f} = a^*_f$, for example, if $\lambda(\{x\in A: f(x) = a\}) = 0$ for any $a\in \mathbb{R}$. In this case $a_f$ can be defined as the unique value for which $\lambda(\{x\in A: f(x) > a\}) = \lambda(A)/2$ (therefore, I called it median).

Turning back to your question, if $f\le g$, then 
$$
\{x\in A: g(x) < a\} \subset \{x\in A: g(x) < a\},\\
  \{x\in A: f(x) \ge  a\}\supset \{x\in A: g(x) \ge  a\}, 
$$
whence it follows that $a_{*,f}\le a_{*,g}$. Similarly, $a^*_{f}\le a^*_{g}$. 
But, naturally, for some particular values $a_f$ and $a_g$ it is possible that $a_f>a_g$. For example, we can take $A=[-1,1]$, $f(x) = 2\operatorname{sign} x$, $g(x) = 2\operatorname{sign} x + 1$, $a_f = 1$, $a_g = 0$.
