Prove that $\int_{I}f=0 \iff$ the function $f\colon I\to \Bbb R$ is identically $0$. Let $I$ be a generalized rectangle in $\Bbb R^n$ 
Suppose that the function $f\colon I\to \Bbb R$ is continuous. Assume that $f(x)\ge 0$, $\forall x \in I$
Prove that $\int_{I}f=0 \iff$ the function $f\colon I\to \Bbb R$ is identically $0$. 

My idea is that
For $(\impliedby)$
Since $f\colon I\to \Bbb R$ is identically zero, $$f(I)=0$$
Then $$\int_{I}f=\int 0=0$$
For $(\implies)$,
Since $f$ is continuous, the function is integrable. 
i.e $\int _{I} f $ exists. 

I need the show that $\int f=0$ but how? 
Hopefully, other solution is true. Please check this. And how to continue this? Thank you:) 

Suppose $f(x)>0$ for some $x\in I$, as $f$ is continuous $\exists\ \delta>0$ such that $\forall y\in B_\delta(x)\cap I$, we will have $f(y)>0$. Then, as $f(x)\ge 0$
$$\displaystyle\int_If\ge\displaystyle\int_{B_\delta(x)\cap I}f>0 \implies\text{contradiction}$$.
