Suppose $f \geq 0$, if continuous on [a,b] and $\int_{a}^{b} f(x)dx=0$. Prove that $f(x)=0$ for all $x\in [a,b]$. My attempt:
Let P the partition of $[a,b]$. Let $x_{i}^{*} \in [x_{i-1},x_{i}]$ and f is non negative in $[x_{i-1},x_{i}]$. 
Since f is continuous on $[a,b]$, then $f$ is R-integrable with $\int_{a}^{b} f(x)dx=0$.
$$\displaystyle \lim_{n\rightarrow \infty} \sum_{i=1}^{n}f(x_{i}^{*}) \delta x_{i}=0$$
This implies $f(x_{i})=0$ for all i. i.e $f(x)=0$.
Is this proof is correct? 

Assume by contradiction that $f(x_0)>0$ for some $x\in [a,b]$. Since $f$ is continuous, then there is a closed ball$^*$ around $x_0$ of radius $\epsilon$ s.t $f(x)>0$ for all points inside this ball. Since the closed ball is compact, $f$ has a (positive) minimum value inside the ball - $m$. What does this tell you about the Lower Darboux sum of $f$ in a partition which involves this ball? Try reaching a contradiction to the fact that the integral of $f$ is $0$.
$*$Note that the assumption about the closed ball is not accurate if $x_0$ is on the boundary of $[a,b]$, but this can easily be fixed - try to think how.
