"The Exercise"
Suppose that $f$ is continuous on $[a,b]$, that $f(x)\ge0,\forall x\in[a,b]$ and that $\displaystyle\int_a^b f=0$. Prove that $f(x)=0$ for all $x\in[a,b]$
My attempt:
We can claim that $f$ has Riemann Integral on $[a,b]$ because of assumption >$\displaystyle\int_a^b f=0$;
so it follows... ;
Since $f$ has Riemann Integral on $[a,b]$, $\forall \epsilon>0$ there exists $\delta>0$ such that if $\dot{P}$ is any tagged partition of $[a,b]$ with $\delta>||\dot P||$ $$\left|S(f;\dot P)-\displaystyle\int_a^b f\right|<\epsilon$$And $$\left|S(f;\dot P)\right|<\epsilon$$ That is $$S(f;\dot P)=\displaystyle\sum_i f(t_i)(x_{i+1}-x_i)=0\quad, \forall i\in\mathbb N\Rightarrow \forall t_i\in[a,b]$$
It is obvious that $(x_{i+1}-x_i)>0$ so $\forall x\in[a,b],\quad f(x)=0$
Why the continuity hypothesis cannot be dropped?I didn't even use it(I think). And if it is, how am I supposed to convince myself (intuitively and logically)?