Second Mean Value Theorem of Integral Proof This is from wikipedia on MVT.

EDIT: Now that user Marsan has given a proof, why does the interval need not contain $a$? $H$ is continuous at $[a,b]$. What happens if $ \int G \phi \, dt = 0$? 

 A: First assume $G$ is a positive decreasing staircase function, say $G= \sum_{i=1}^n f_i \cdot \chi_{]x_{i-1}, x_i[}$, where $a= x_0 < x_1 < \cdots<x_n=b$, $f_i$ are positive real numbers and $\chi_{]x_{i-1}, x_i[}$ denotes the indicator function of the interval $]x_{i-1}, x_i [$. Let $H: x \in [a,b] \mapsto \int_a^x \phi $. The function H is (absolutely) continuous. By the intermediate value theorem it suffices to show that $$ m G(a^+) \leq \int_a^b G\phi \leq M G(a^+)$$ where $ m = \min H$ and $M = \max H$. Notice that $f_1 = G(a^+)$. Now, \begin{align}\int_a^b G\phi &= \sum_{i=1}^n \int_{x_{i-1}}^{x_{I}}f_i\phi \\ & = \sum_{i=1}^n f_i (H(x_{i}) - H(x_{i-1})) \\ & = \sum_{i=1}^n f_i H(x_i) - \sum_{i=1}^n f_i H(x_{i-1})\\ & = \sum_{i=1}^{n-1}\underbrace{(f_i - f_{i+1})}_{\geq 0}H(x_i) + f_nH(b)\end{align} This yields $$ m\left(\sum_{i=1}^{n-1}(f_i -f_{i+1}) + f_n \right) \leq \int_a^b G\phi \leq M\left(\sum_{i=1}^{n-1}(f_i -f_{i+1}) + f_n \right) $$ and thus we get the inequality which was desired, which proves the result. 
Now you can easily approximate $G$ as a limit of staircase positive decreasing functions to conclude. 
A: For a different approach, the function  $\varPhi: x \mapsto  \int_a^x \varphi(t) \, dt$ is continuous and since $G$ is monotone we have  that $\varPhi$ is Riemann-Stieltjes integrable with respect to $G$.  From integration by parts, it follows that $G$ is integrable with respect to $\varPhi$ and
$$\int_a^b G \, d\varPhi= G(b)\varPhi(b) - G(a)\varPhi(a) - \int_a^b \varPhi \, dG  = G(b)\varPhi(b) - \int_a^b \varPhi \, dG.$$
It also follows from a general property of the Riemann-Stieltjes integral that if $G$ and $\varphi$ are Riemann integrable, which is the case here, then
$$\int_a^bG(t) \varphi(t) \, dt = \int_a^b G \, d\varPhi = G(b)\varPhi(b) - \int_a^b \varPhi \, dG.$$
Since $\varPhi$ is continuous, by the first mean value theorem for integrals there exists $x \in [a,b]$ such that 
$$\int_a^b \varPhi \, dG = \varPhi(x)\int_a^b dG = \varPhi(x)[G(b) - G(a)].$$
Hence,
$$\begin{align}\int_a^bG(t) \varphi(t) \, dt &= G(b)\varPhi(b) - \varPhi(x)[G(b) - G(a)] \\ &= G(a) \varPhi(x) + G(b)[\varPhi(b) - \varPhi(x)] \\ &= G(a) \int_a^x \varphi(t) dt + G(b) \int_x^b \varphi(t) \, dt \end{align}$$
Since we can obtain the same result by replacing $G$ with $\hat{G}$ where $\hat{G}(t) = G(t) $ on $[a,b)$ and $\hat{G}(b) = 0$ we obtain 
$$\begin{align}\int_a^bG(t) \varphi(t) \, dt &= \int_a^b\hat{G}(t) \varphi(t) \, dt  \\ &= \hat{G}(a) \int_a^x \varphi(t) dt + \hat{G}(b) \int_x^b \varphi(t) \, dt \\ &= G(a) \int_a^x \varphi(t) dt \end{align} $$
Note that there is no reason why we must restrict $x \in (a,b]$. This would become evident by working through the proof of the first mean value theorem for integrals where no such restriction applies. If $\varPhi$ is continuous, attaining minimum and maximum values at $\xi_1$ and $\xi_2,$ and it happens that $\varPhi(\xi_1) < \varPhi(a) < \varPhi(\xi_2)$ then $\varPhi(a) = \varPhi(x)$ for at least one other point in $(a,b)$. It is possible that this is a point where $\varPhi(x) = \int_a^b \varPhi \, dG / [G(b) - G(a)]$.
