Proving that if the Lower (Darboux) Integral $L(f) = 0$ then $f(x) = 0$ I'm having trouble proving this question from my textbook regarding Darboux integrals and was hoping for some pointers.

Let $f$ be continuous on $I:=[a,b]$ and assume $f(x)\ge0$ for all $x \in I$. Prove if $L(f) = 0$, then $f(x) = 0$ for all $x \in I$.

My approach so far has been pretty direct but I'm a bit stuck.
Proof: 
Suppose $L(f)=0$ for $f$ continuous on $I:=[a,b]$ and $f(x)\ge0$.
Then $$sup\{L(f,P): P \in \wp\} = 0$$ So we can write $$L(f,P) = \sum\limits_{k=1}^n m_k ({x_{k-1} - x_k}) \le 0$$ Now $(x_{k-1} - x_k) > 0$ for all $k=1,2,...,n$ so then $m_k \le 0$. Write $$m_k = inf \{ f(x): x \in [x_{k-1}, x_k] \} \le 0$$
So for all $x \in [x_{k-1}, x_k]$, $$0 \le f(x) \le 0$$
So this is where I'm stuck, as I'm not sure if my last step is correct and how to show that it holds for all $x \in I$.
 A: Suppose not.  If $f \neq 0$ at some $x \in I$, then there exists an $\epsilon > 0$ such that  $f \ge \epsilon$ on some small interval $(a,b) \subset I$ (why?).  Now you should be able to derive a contradiction.
A: Or we can solve it by another method. Since, $f(x)$ is continuous in a closed and bounded interval so f is riemann integrable. So, $\int_a^bf(x)dx=L(f)=0\ \forall x\in [a,b].$ Now here the analysis comes, that $\int_a^bf(x)dx=0 $ and $f(x)\geq 0\ \forall x\in [a,b], \implies f(x)=0\ \forall\ x\in [a,b].$ For this we will use the contradiction that Let $f(x)\neq 0\ i.e. f(x)>0\ \forall\ x\in[a,b].$ So,if we can prove that " If a continuous function is not zero at one point then there exists some open interval such that that function is non-zero, then we are done because in that open interval the functional value is strictly greater than 0 , so the integration is not going to be zero. "  For proving the last argument we can use the definition of continuity. As $f(x)$ is continuous at $x_0; x_0\in(a,b)$ $\forall \ x\in [a,b] \implies \forall\ \epsilon>0\ \exists\ \delta >0\ \text{s.t}\ \forall\ x\in(x-\delta,x+\delta) \ f(x) \in (f(x)-\epsilon,f(x)+\epsilon). $  Choose $\epsilon=\frac{|f(x)|}{2}$ so, $\exists\ \delta$ s.t. $\forall\ x\in(x_0-\delta,x_0+\delta), f(x)\in(\frac{|f(x_0)|}{2},\frac{3|f(x_0)|}{2}).$ Hence we got an interval $x_0-\delta,x_0+\delta)$ s.t. $f(x)>0, $ which contradicts that the integral is zero. Therefore, we proved that $f(x)$ must be equal to zero.
