If $ f > 0 $, $g\ge0$, and $ \int_a^b g > 0 $, then $ \int_a^b fg > 0 $? Let $ f:[a,b] \rightarrow \mathbb{R} $ be continuous on $ [a,b] $ and $ f > 0 $ on $ (a,b) $, and let $ g:[a,b] \rightarrow \mathbb{R} $ be non-negative and integrable on $ [a,b] $. 
If $ f > 0 $, and $$ \int_a^b g > 0 \,, $$ then is it true that $$ \int_a^b fg > 0 \,? $$ 
The integrals are Riemann integrals. 
 A: This answer uses Riemann integrals.
If $g$ is non-negative and has a positive integral then it is positive in some subinterval which implies that $fg$ is non-negative and positive in some sub-interval. Thus it has a positive integral. Continuity of $f$ is not needed. Rather the following non-trivial theorem is needed :

Theorem: If $f$ is Riemann integrable on $[a, b] $ with a positive integral then $f $ is positive on some sub-interval of $[a, b] $. 

A: An extension to the Mean Value Theorem for Integrals is 

Proposition. If $ f,g $ are continuous on $ [a,b] $ and $ g(x) > 0 $, for all $ x \in [a,b] $, then there exists $ c \in (a,b) $ such that $$ \int_a^b fg = f(c) \int_a^b g \,. $$

which is taken from Exercise 17 of Section 7.2 from Bartle and Sherbert's Introduction to Real Analysis. 
Applying this gives the desired result, since $ f > 0 $. 
A: If $\epsilon>0$ is small enough, then $\int_{a+\epsilon}^{b-\epsilon}g>0$.  Moreover, $f\ge C(\epsilon)>0$ on $[a+\epsilon,b-\epsilon]$, so 
$$
\int_a^b fg\ge \int_{a+\epsilon}^{b-\epsilon}fg\ge C(\epsilon)\int_{a+\epsilon}^{b-\epsilon}g,
$$
which is positive.
