Q:

Does $f \geq 0 \forall x$ and $f>0$ for a countable set imply $\int_a^b f >0$?

Let $f:\Bbb R \to \Bbb R$ be continuous.
Does $f \geq 0 \forall x$ and $f>0$ for a countable set of points $A$ imply $\int_a^b f(x) dx >0$ if $A\subseteq (a,b)$?
I've had this doubt. I'm fairly sure this is not true, but I think it would take a pretty weird function to disprove it.
Does anyone know a simple proof/counterexample?

A:

If $f(x)$ is continuous, $f(x)\geq 0$ and $f(x)>0$ for a SINGLE point, then $\int_a^b f(x) dx>0$
Use continuity to get a small interval around the SINGLE point $x_0$, where $f(x)>\frac{f(x_0)}{2}$ using the $\epsilon-\delta$ definition of a limit. 
Edit, see for example $f\geq 0$, continuous and $\int_a^b f=0$ implies $f=0$ everywhere on $[a,b]$

