Well, I was trying to prove this, particularly $(2)\Rightarrow (3)$ without using more than the definition of being Riemann integrable (not using density of continuity points or Lebesgue measure 0 for the set of points in which f is not continuous):
Let $f:[a,b]\to\mathbb{R}$ a Riemann integrable function. The following statements are equivalent:
$(1)$ $\int^{b}_{a}|f(x)|dx=0$
$(2)$ If $f$ is continuous at $c \in [a,b]$, then $f(c)=0$
$(3)$ The interior of $X=\{x\in [a,b]: f(x)\neq 0\}$ is empty.
Can someone help me?
(2) $\implies$ (3)
Assume that $(3)$ is false and $X^o$, the interior of $X$, is non-empty. Then there is a non-empty open interval $I \subset X^o$ such that $|f(c)| > 0$ for every $c \in I$. If $f$ is continuous at any $c \in I$ contradicting condition $(2)$. Otherwise, $f$ must be discontinuous everywhere in $I$ which violates the premise that $f$ is Riemann integrable. Thus if condition $(3)$ is false, so too is condition $(2)$, and $(2) \implies (3).$
(1) $\implies$ (3)
Suppose $X^o$, the interior of $X$, is non-empty. Since $X^o$ is an open set it is a countable union of disjoint non-empty open intervals where $|f(x)| > 0$
$$X^o = \bigcup_{k=1}^\infty (a_k,b_k),$$
and for some finite $n$ we get the contradiction
$$\int_a^b |f(x)| \, dx \geqslant \sum_{k=1}^n \int_{a_k}^{b_k}|f(x)| \,dx > 0.$$
(1) $\implies$ (2)
Suppose $f(c) \neq 0$ and $f$ is continuous at $c$. Then there exists an interval $(c - \delta, c+ \delta) \subset [a,b]$ where $|f(x)| > |f(c)|/2 > 0$ and we get the contradiction
$$\int_a^b |f(x)| \, dx \geqslant \int_{c-\delta}^{c + \delta}|f(x)| \, dx > 2 \delta \frac{|f(c)|}{2} > 0.$$
This should help you get started.