Answer
False.
Work Step by Step
Just because the average value of $g(x)$ is less than the average value of $f(x)$ doesn't mean that $g(x)$ can't be greater than $f(x)$ at a single point. For example, the average value of $f(x)=2$ on the interval $[0,3]$ is greater than the average value of $g(x)=x$ on $[0,3]$, but $g(3)$ $(y=3)$ is greater than $f(3)$ $(y=2).$
