If $\int f=\int g$ and $\int_0^t f \geq \int_0^t g$ for all $t≥0$, then is $\int_0^t g^{-1} \geq \int_0^t f^{-1}$ for all $t\geq 0$? Suppose $f,g$ are continuous, integrable, decreasing, nonnegative-real-valued functions, each defined on some interval of nonnegative real numbers with left endpoint $0$, and satisfying $\inf f = \inf g = 0$. (The intervals can be open, half-open, or closed, and may be infinite.) The assumptions imply that $f$ and $g$ have similarly continuous, integrable, decreasing inverses (I mean inverse with respect to function composition) defined on similar intervals. Then suppose
$$\int f = \int g,$$
where the integration is over their domains. Finally, suppose that for all $t$ for which both functions are defined, we have
$$ \int_0^t f \geq \int_0^t g.$$
It seems to me that in this situation it follows that
$$ \int_0^t g^{-1} \geq \int_0^t f^{-1} $$
for all $t$ for which both $f^{-1}$ and $g^{-1}$ are defined, as well. But I think this for an essentially hand-wavy reason: "Since the functions are decreasing and trapped in the first quadrant, then if mass is moved downard, it has to go to the right."
Question: Is the claim correct, and if so, how is it properly proven?
 A: Here we only cosider the case where $f$ and are define in the bounded interval $[0,a]$,   and thus $f(a)=0=g(a)$; we further assume that  $f(0)=g(0)=b$.
Hence, as  $f$ and $g$ are strictly monotone decreasing continuous on $[0,a]$,  so are $f^{-1}$ and $g^{-1}$ (on $[0,b]$).
Since $\int^b_0f^{-1}=\int^a_0f=\int^a_0g=\int^b_0g^{-1}$ and $\int^t_0f\geq \int^t_0g$ for all $0\leq t<a$, $f-g$ and $f^{-1}-g^{-1}$ admit changes of sign in $[0,a[$ and $[0,b]$ respectively.
Suppose there is $0<s<g(0)$ such that $\int^s_0f^{-1}>\int^s_0g^{-1}$. If $f^{-1}(s)=g^{-1}(s)$, set $S=s$; if $f^{-1}(s)>g^{-1}(s)$, there is $S>s$ such that $f^{-1}(S)=g^{-1}(S)$ and $\int^S_0f^{-1}>\int^S_0g^{-1}$, otherwise $f^{-1}(v)>g^{-1}(v)$ for all $s\leq v< b$ and so, $\int^b_0f^{-1}>\int^b_0g^{-1}$ which is a contradiction; If $f^{-1}(s)<g^{-1}(s)$, there is $0<S<s$ such that $f^{-1}(S)=g^{-1}(S)$ and $\int^S_0f^{-1}>\int^S_0g^{-1}$, otherwise $f^{-1}(v)<g^{-1}(v)$ for all $0<v<s$ and so, $\int^s_0f^{-1}<\int^s_0g^{-1}$ which is a contradiction. Set $T=f^{-1}(S)$. Then
\begin{align}
\int^a_0f&=\int^T_0f+\int^a_Tf=\int^T_0 f-ST+\int^S_0 f^{-1}\\
&>\int^T_0g - ST+\int^S_0g^{-1}=\int^T_0g+\int^a_Tg=\int^a_0g
\end{align}
which is not possible.
A: Yes, the claim is correct.
Without loss of generality we can assume that the domains of $f$ and $g$ are open (possibly infinite) intervals.
We consider the function $f$ first: Let the domain of $f$ be $(0, a)$ for some $0 < a \le \infty$. $f$ is continuous, strictly decreasing, with
$$
 \inf \{ f(x) \mid 0 < x < a \} = \lim_{x \to a-} f(x) = 0 \, .
$$
The image of $f$ is the interval $(0, b)$ where
$$
 b = \sup \{ f(x) \mid 0 < x < a \} = \lim_{x \to 0+} f(x)
$$
satisfies $0 < b \le \infty$. The inverse function $f^{-1}$ maps $(0, b)$ to $(0, a)$. We also know that $f$ is integrable, i.e. $I = \int_0^a f(x) \, dx $ exists and is finite.
Now let $c \in (0, a)$. By applying Fubini's theorem to the measure of the set
$$
 \{ (x, y) \mid 0 < x < a, \, 0 < y < \min(f(c), f(x))\} \\
 = \{ (x, y) \mid 0 < y < f(c), \, 0 < x < f^{-1}(y)\} 
$$
one can see that
$$ \tag{*}
 \int_0^{f(c)} f^{-1}(y) \, dy = c f(c) + \int_c^a f(x) \, dx
= c f(c) + I - \int_0^c f(x) , dx \, .
$$
Taking the limit $c \to 0$ we get that in particular
$$
 \int_0^b f^{-1}(y) \, dy = \int_0^a f(x) \, dx  = I\, .
$$
(This is true even if $f$ is unbounded, the monotonicity and integrability of $f$ implies that $\lim_{c \to 0} cf(c) = 0$.)
Similarly, $g$ is a 1-1 mapping from some interval $(0, a')$ to $(0, b')$, where again $a', b'$ are positive and can be $+\infty$. Also
$$ \tag{**}
\int_0^{g(c)} g^{-1}(y) \, dy =  c g(c) + I - \int_0^c g(x) , dx \, .
$$
for $0 < c < a'$, since $\int_0^{a'} g(x)dx = I = \int_0^a f(x) dx$, and in particular
$$
 \int_0^{b'} g^{-1}(y) \, dy = \int_0^{a'} g(x) \, dx  = I\, .
$$
Now we are ready to show that
$$
 H(s) = \int_0^s g^{-1}(y) \, dy - \int_0^s f^{-1}(y) \, dy
$$
is nonnegative for $0 \le s \le \min(b, b')$.
If $b < b'$ then
$$
 \int_0^t (f(x) - g(x))\, dx < 0
$$
for sufficiently small $t$, in contrast to the assumption. So this is not possible, and we have $\min(b, b') = b'$.
$H$ is continuous on $[0, b']$ and differentiable on $(0, b')$, with $H(0) = 0$ and
$$
 H(b') = \underbrace{\int_0^{b'} g^{-1}(y) \, dy
 - \int_0^{b} f^{-1}(y) \, dy }_{= I - I = 0}
 +\int_{b'}^b f^{-1}(y) \, dy \ge 0 \, .
$$
If $H$ attains its minimum at some some point $d$ in the interior of the interval then
$$
0 = H'(d) = g^{-1}(d) - f^{-1}(d)
$$
and using $(*)$ and $(**)$ with $c = f^{-1}(d) = g^{-1}(d)$ we get
$$
 \int_0^{d} f^{-1}(y) \, dy = cd + I - \int_0^c f(x) \, dx \\
\le cd + I - \int_0^c g(x) \, dx =  \int_0^{d} g^{-1}(y) \, dy
$$
and therefore $H(d) \ge 0$.
We have thus shown that $H(s) \ge 0$ everywhere on $[0, b']$, and that concludes the proof.
