Demonstration of $\int_{a}^b f(x) \,dx= 0 \Rightarrow f(x)\equiv0 $ Good morning,

Can you give me a help to demonstrate this proposition:

$f$ is a continuous and not negative function on the interval $[a,b] \ a,b \in \Re $,  Demonstrate:

$$\int_{a}^b f(x) \,dx= 0 \Rightarrow f(x)\equiv0  $$
Let 

$$\Phi(t)=\int_a^tf(x)dx$$
then $\Phi'(t)=f(t)\ge0$ hence $\Phi$ is non-decreasing but since $\Phi(a)=\Phi(b)=0$, this means $\Phi$ is constant and $\Phi'(t)=f(t)=0$.
If $f$ is Riemann integrable on $[a,b]$, then $f$ is Lebesgue integrable on $[a,b]$, and the Riemann and Lebesgue integrals are equal. It is fairly straightforward to show that a nonnegative measurable function has Lebesgue integral zero if and only if it is zero almost everywhere. Since we have the additional assumption that $f$ is continuous, if $f$ is zero almost everywhere, then necessarily $f$ must be identically zero.

I don't understand why I have 3 down votes for this answer and no comments...