I was trying to prove this three result which are equivalent. Recall $f$ is said to have the Intermediate Value Property in $[a, b]$ if $\forall c,d \in [a, b], \ y \in (f(c), f(d)) \ \exists x \in (c, d) \ / \ f(x)=y $
Statement 1
Let $f:[a, b] \rightarrow\mathbb{R}$, with $f \geq 0$. Suppose that $f$ has the Intermediate Value Property. Suppose also that $f \in \mathcal R[a, b]$ and that $\int_a^b f(x)\,dx = 0$. Then $f(x)=0, \ \forall x \in [a, b]$.
Statement 2
Let $f:[a, b] \rightarrow\mathbb{R}$. Suppose that $f^2$ has the Intermediate Value Property. Suppose also that $f^2 \in \mathcal R[a, b]$ and that $\int_a^b f^2(x)\,dx = 0$. Then $f(x)=0, \ \forall x \in [a, b]$.
Statement 3
Let $f:[a, b] \rightarrow\mathbb{R}$. Suppose that $f$ has the Intermediate Value Property. Suppose also that $f^2 \in \mathcal R[a, b]$ and that $\int_a^b f^2(x)\,dx = 0$. Then $f(x)=0, \ \forall x \in [a, b]$.
$\boxed{1 \Rightarrow 2}$
Trivial, as $f^2$ satisfies every condition for Statement 1, so $f^2 = 0$ and then $f = 0$.
$\boxed{2 \Rightarrow 3}$
If $f$ has the Intermediate Value Property, so does $f^2$ (Not very hard to prove)
$\boxed{3 \Rightarrow 1}$
As $f$ is bounded because of being integrable, $\exists K \geq f(x); K\geq 0$. Also, as $f \in \mathcal R[a, b]$, $f^2 \in \mathcal R[a, b]$ holds. Then
$0 \leq f^2 \leq Kf \Rightarrow 0 = \int_a^b 0 dx \leq \int_a^b f^2(x) dx \leq K \int_a^b f(x) dx = 0 \Rightarrow \int_a^b f^2(x) dx = 0$, and Statement 3 holds
The three statement are generalizations of weaker results. If we assume continuity or having a primitive the result holds. I would like to have an "elementary" proof without Measure Theory or Lebesgue Integral.
Update: The answer is completely revamped with an easier counter-example.
The statement is actually false.
Recall that a function satisfying the Intermediate Value Property is called a Darboux function.
Let $g : [0, 1] \to [0, 1]$ be any Darboux function which is not identically zero and vanishes on any dyadic rationals in $[0, 1]$ (that is, $g(k/2^n) = 0$ for all $n \geq 0$ and $0 \leq k \leq 2^n$.)
For example, such a function can be found by imitating the construction of Conway base 13 function.
Let $\phi : [0, 1] \to [0, 1]$ be the Cantor function. Since $\phi$ is continuous on $[0, 1]$, it is a Darboux function.
Then the composition $f = g \circ \phi$ satisfies the following properties:
If $C$ is the Cantor set, then $\phi([0,1]\setminus C)$ only consists of dyadic rationals, and so, $f$ is identically zero outside of $C$.
Since $f$ is the composition of two Darboux functions, it is again a Darboux function.
Since $0 \leq f \leq \mathbf{1}_C$, where $\mathbf{1}_C$ is the indicator function for $C$, and $\mathbf{1}_C$ is Riemann-integrable with $\int_{0}^{1} \mathbf{1}_C = 0$, it follows that $f$ is also Riemann integrable and $\int_{0}^{1} f = 0$.
Therefore $f$ provides a counter-example to Statement 1.