Proof of a theorem of Bonnet about integrals In this paper from 1849, Bonnet claims that it's "very easy" to prove the following statement

Let $f,\phi:[a,b]\to \mathbb R$ be two continuous functions. Suppose $\forall x\in [a,b], A\leq \int_a^x \phi(t) dt\leq B$. If $f\geq 0$ and $f$ decreases then $\forall x\in [a,b], Af(a)\leq \int_a^x \phi(t)f(t) dt\leq Bf(a)$

Can you prove Bonnet's statement ? If $\phi \geq 0$, it's not difficult to prove $\int_a^x \phi(t)f(t) dt\leq Bf(a)$, but I can't find anything else...
 A: This is generally true with $\phi$ integrable and $f$ decreasing.  One approach is to use Riemann sums for proof.  Another is to use Riemann-Stieltjes integration.  
Defining $\Phi(x) = \int_a^x \phi(t) \, dt$ we have $\Phi$ absolutely continuous, and it is a general property of Riemann-Stieltjes integrals that (with $f$ monotone),
$$\int_a^b \phi(x)f(x) \, dx = \int_a^b f \, d\Phi. $$ 
Applying integration by parts we obtain
$$\int_a^b \phi(x)f(x) \, dx = \int_a^b f \, d\Phi = \Phi(b)f(b) - \Phi(a)f(a) - \int_a^b \Phi \, df.$$
Since $\Phi(a) = 0$ it follows that
$$\tag{*}\int_a^b \phi(x)f(x) \, dx =  \Phi(b)f(b) - \int_a^b\Phi \,df.$$
Since $\Phi$ is continuous it is bounded on $[a,b]$.  Let $A = \inf_{x \in [a,b]}\Phi(x)$ and $B = \sup_{x \in [a,b]}\Phi(x)$.
Since $A \leqslant \Phi(x) \leqslant B$ for all $x \in [a,b]$ and $f$ is decreasing we have
$$\tag{**}A(f(a) - f(b)) \leqslant -\int_a^b \Phi\, df \leqslant B(f(a) - f(b)).$$
From (*) and (**) it follows that
$$Af(a) - Af(b) + \Phi(b) f(b) \leqslant \int_a^b \phi(x) f(x) \, dx \leqslant Bf(a) - Bf(b) + \Phi(b) f(b),$$
and
$$Af(a) + (\Phi(b) - A)f(b) \leqslant \int_a^b f(x) h(x) \, dx \leqslant Bf(a) -(B- \Phi(b))f(b).$$
Since $A \leqslant \Phi(b) \leqslant B$ and $f$ is non-negative, we have $(\Phi(b)-A)f(b) \geqslant 0$ and $(B- \Phi(b))f(b) \geqslant 0$.
Therefore,
$$Af(a) \leqslant \int_a^b \phi(x) f(x) \, dx \leqslant Bf(a).$$
