Prove a version of the mean value theorem where f is decreasing I'm trying to prove that if $f$ is a non-negative and decreasing function, and $h$ is continuous on a $[a,b]$, there there exists a point $c\in[a,b]$ such that 
$$\int_a^b f h = f(a)\int_a^c h$$
I know that the result is valid if we suppose that $f$ is increasing instead of assuming that is decreasing. So I guess the result that I'm trying to prove is almost there, but I can't see it. 

First define
$$H(x) = \int_a^xh(t)\,dt.$$
Since $h$ is a continuous function, it follows that $H$ is differentiable and $f$ is Riemann-Stieltjes integrable with respect to $H$
$$\int_a^bf(x)h(x) \,dx= \int_a^bf(x)H'(x) \,dx=\int_a^bf \, dH.$$
Integrating by parts,
$$\int_a^bf(x)h(x) \,dx= f(b)H(b)-f(a)H(a)-\int_a^bH \, df.$$
By the mean value theorem for integrals, there is a point $c \in [a,b]$ such that
$$\int_a^bf(x)h(x) \,dx= f(b)H(b)-f(a)H(a)-H(c)[f(b) - f(a)] \\=f(a)[H(c)-H(a)]+f(b)[H(b)-H(c)]\\=f(a)\int_a^{c}h(x) \, dx + f(b)\int_{c}^{b}h(x) \, dx.$$
Since $f$ is non-negative and decreasing, we can repeat the above derivation for the non-negative, decreasing function $\hat{f}$ given by 
$$\hat{f}(x) = \begin{cases}f(x),  \,\,x \in [a,b) \\ 0, \,\,\,\,\,\,\,\,\,\,\,x = b\end{cases}$$ 
Hence,
$$\int_a^bf(x)h(x) \,dx = \int_a^b\hat{f}(x)h(x) \,dx=f(a)\int_a^{c}h(x) \, dx.$$
The integrals involving $f$ and $\hat{f}$ are equal since changing the value of a function at a finite number of points does not change the value of a Riemann integral.
