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:
$$\int_{a}^b f(x) \,dx= 0 \Rightarrow f(x)\equiv0  $$ 
 A: 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 $\Phi'(t)=f(t)=0$.
A: Suppose that $f(x) = \varepsilon 0$ for some $x \in [a,b]$. Because $f$ is continuous there exists $\delta$, that $f(y)=\frac{\varepsilon}{2}>0$ for $y \in [x-\delta,x+\delta]$, so:
$$\int_{a}^{b}f(x)dx=\int_{a}^{x-\delta}f(x)dx+\int_{x-\delta}^{x+\delta}f(x)dx+\int_{x+\delta}^{b}f(x)dx$$
Now $\int_{a}^{x-\delta}f(x)dx\geq0$ and $\int_{x+\delta}^{b}f(x)dx \geq 0$ because $f$ is non-negative, so:
$$\int_{a}^{b}f(x)dx \geq \int_{x-\delta}^{x+\delta}f(x)dx$$
but $$f(y)=\frac{\varepsilon}{2}>0$$ for $y \in [x-\delta,x+\delta]$
so
$$\int_{a}^{b}f(x)dx \geq \int_{x-\delta}^{x+\delta}f(x)dx>\frac{\varepsilon}{2}\cdot 2\delta >0$$
A: 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... 
