Second Mean Value Theorem

Question

Suppose $f:[a, b] \to R$ is a positive decreasing function Integrable on $[a, b]$ and $g(x)$ is Integrable on $[a, b]$ then prove that $$\int_a^bf(t)g(t)dt=f(a^+)\int_a^cg(t)dt$$, for some $a \leq c < b$ where $f(a^+)=\lim_{x \to a^+}f(x)$. $$$$As $f(x)$ is decreasing on $[a, b]$ so $$f(b) \leq f(x) \leq f(a)$$ for all $x$ in $[a, b]$.Now as $f(x)$ is bounded from above so it must have some least Upper Bound say $L$.Now consider the interval $(a, a+\delta)$ for some $\delta>0$ then in this interval $f(x)$ is decreasing and has the least Upper Bound $L$ so we get $$\lim_{x \to a^+}f(x)=L$$. So we have $$0<f(x) \leq L$$ for all $x$ in $[a, b]$. And hence we get $$\frac{|\int_a^bf(t)g(t)dt|}{L} \leq \frac{\int_a^b|f(t)||g(t)|dt}{L} \leq \frac{L\int_a^b|g(t)|dt}{L} = \int_a^b|g(t)|dt$$. Now as the function $\int_a^xg(t)dt$ is continuous on $[a, b]$. So if we can prove the inequality $$\text{min}(\int_a^xg(t)dt) \leq \int_a^b|g(t)|dt \leq \text{max}(\int_a^xg(t)dt)$$ then by intermediate value theorem we can prove required thing but I am unable to prove the inequality. How to prove this??