If $h$ is closer to $f$ than to $g$, its integral on $\{f > g\}$ must "agree" with $f$'s? I have the following question (a result I would like to prove, with an admirable record of failing at it so far) -- I actually do not know if it is obvious or just plain wrong.
Let $f,g,h\colon [0,1]\to\mathbb{R}_+$ be 3 (integrable) functions such that $\int_{[0,1]}f=\int_{[0,1]}g=\int_{[0,1]}h=1$; set $$A_{fg}\stackrel{\rm def}{=}\left\{x\in[0,1]\mid f(x) > g(x)\right\}$$
and write $\omega_f=\int_{A_{fg}} f$, $\omega_g=\int_{A_{fg}} g$, $\omega_h=\int_{A_{fg}} h$.
Suppose $\int_{[0,1]} \lvert g - h\rvert \geq \int_{[0,1]} \lvert f - h\rvert + \gamma$, for some constant $\gamma > 0$. Can we prove a quantitative bound on $\omega_h$, namely that it must be "closer to $\omega_f$ than to $\omega_g$?".
I hope that a statement of this form holds:

There exists an absolute constant $c>0$ (e.g., $1/10$) such that $\omega_h > \frac{\omega_f+\omega_g}{2}+c\cdot\gamma$.

It does sound "intuitive" to me, yet I have learnt not to give too much credit to my intuition. Especially since I have been struggling with this for one day now, and quite a few sheets of paper: I know how to prove it with the extra assumption that $\int_{[0,1]} \lvert f - h\rvert \leq \frac{\gamma}{10}$, but this is not enough for my purpose.
Small note: one can assume without loss of generality $f,g,h$ to be "regular enough" (continuous, for instance).
 A: This will not work. For small $\delta>0$, let's define $f(x)=1$,
$$
g(x) = \begin{cases} 0 & 0<x<1/2 \\ 1 & 1/2 < x < 1-\delta \\
\frac{1+2\delta}{2\delta} & 1-\delta<x< 1 \end{cases} , \quad
h(x) = \begin{cases} \frac{1+\delta}{2\delta} & 1/2< x < 1/2 + \delta \\
1/2 & \textrm{otherwise} \end{cases} .
$$
Then $A=[0,1/2]$, so $\omega_h=(1/2)(\omega_f+\omega_g)$ (since $g=0$, $h=1/2$, $f=1$ on $A$), and I have a counterexample if I can verify that $\|g-h\|_1>\|f-h\|_1$.
This we can of course do by a straightforward explicit calculation. However, we can also (more conveniently) argue as follows: Observe that the bumps of $g$, $h$ have area $\approx 1/2$ (they have to, to make $\int f=\int g=1$). Now away from the bumps, that is, for most points in $[0,1]$, we have that $|f-h|=|g-h|=1/2$. We then observe that $\int |g-h|$ picks up two bumps, while only the bump of $h$ contributes to $\int |f-h|$. Thus $\int|f-h|\approx 1$, $\int |g-h|\approx 3/2$, and in particular, $\int|g-h|>\int |f-h|$, as desired.
