Proving that if the Lower (Darboux) Integral $L(f) = 0$ then $f(x) = 0$

Question

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 $x \in I$.

Answer 1

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.