$f(x) \ge 0$ for all $x \in [a, b]$ and $\int_a^b f = 0$. Prove that $f(x) = 0$ for all $x \in [a, b]$.

Suppose that $f$ is continuous on $[a, b]$, that $f(x) \ge 0$ for all $x \in [a, b]$ and that $\int_a^b f = 0$. Prove that $f(x) = 0$ for all $x \in [a, b]$.

My attempt:

Let $\dot{\Pi}$ be a tagged partition of $[a, b]$. We have that for any $\varepsilon > 0$ there is some $\delta_\varepsilon > 0$ such that $|\dot{\Pi}| < \delta_\varepsilon$ and$|S(f, {\dot{\Pi}})| = |\sum_{i} f(t_i)(x_{i}-x_{i-1})| < \varepsilon$.

Assume $f(t_i) \neq 0$ for some $i$. Since we know that $x_i - x_{i-1} > 0$ (by the definition of the tagged partition) and $f(t_i) > 0$ then $f(t_i)(x_i-x_{i-1}) > 0$. Then $S(f, {\dot{\Pi}}) > 0$. By Archimedian property there is some $m \in \mathbb{N}$ such that $\frac{1}{m } < S(f, {\dot{\Pi}})$. Lets take $\varepsilon = \frac{1}{m}$. Then $|S(f, {\dot{\Pi}})| \ge \varepsilon$ i.e. $f \not\in \mathcal{R}[a, b]$.So, if $S(f, {\dot{\Pi}}) = 0$ we should have $f(t_i) = 0$ for all tags $t_i$. Since ${\dot{\Pi}}$ was arbitrary and the tags were arbitrary then $f(x) = 0$ for all $x \in [a, b]$.

What is wrong with this logic? I know something is off since the next thing for me is to show that the proposition is not valid unless $f$ is continuous and I never used the continuity of $f$ in my proof.

An alternative - with the usage of the continuity of $f$ (again, seems to come to the same point and getting stuck)

Assume that $f$ is not identically $0$. Then there is some $c \in [a, b]$ such that $f(c) > 0$ for some $c \in [a, b]$. Since $f$ is a continuous function then there is some $\delta > 0$ such that if $|x-c| < \delta$ then $|f(x) - f(c)| < \frac{f(c)}{2}$. So, $f(x) > f(c) / 2$ for $x \in [c-\delta, c + \delta] \cap [a, b]$.Let ${\dot{\Pi}}$ be the tagger partition of $[a, b]$ that includes the sub-interval $[c-\delta, c + \delta] \cap [a, b]$. Let's denote that sub-interval $[x_{i-1}, x_{i}]$. Note that for any tag from this sub-interval we have $f(t)(x_i-x_{i-1}) > \frac{f(c)}{2}(x_i - x_{i-1})$.

Since $x_i > x_{i-1}$ then $f(t)(x_i-x_{i-1}) > 0$ and $S(f, \dot{\Pi}) > 0$.At this point I would like to use the Archimedian again. However, people say that it is a circular logic. I just don't know...