Q:

A problem in continuity of f(x)

(Apostol 3.20.7) Suppose $f$ is integrable and non-negative on $[a, b]$, and that $\int_a^b f(x) dx = 0$. Show that $f(x) = 0$ at each point of continuity of $f$.

A:

Suppose $\exists\ x$ such that $f$ is continuous at $x$ and $f(x)=c\neq 0$. Then due to continuity of $f$ at $x$ we have that for some $\delta$, in $(x-\delta,x+\delta)$, $f(x)\in (c/2,3c/2)$, where we are taking $\epsilon = c/2>0$ in the definition of continuity.
Therefore 
$ \int_a^b f(x)\,dx.$ = $ \int_a^{x-\delta} f(x)\,dx.$ + $ \int_{x-\delta}^{x+\delta} f(x)\,dx.$  + $ \int_{x+\delta}^b f(x)\,dx.$.
Task for you :- deduce that $ \int_a^b f(x)\,dx.\neq 0$ and arrive at the desired contradiction.

