Let $f$ be continuous on $[a,b]$ and suppose that $f(x) \ge 0$, $ \forall x \in [a,b]$. Prove that if $L(f) = 0$ then $f(x) = 0$

Question

Let $f$ be continuous on $[a,b]$ and suppose that $f(x) \ge 0$, $ \forall x \in [a,b]$. Prove that if $L(f) = 0$ then $f(x) = 0, \space \forall x \in [a,b]$.

Quick definitions in case you haven't seen these:

$L(f,P)= \sum_{i=1}^n m_i(f) \Delta x_i$

$m_i(f) = inf\{f(x) |\space x \in [x_{i-1},x_i]\}$

$L(f) = sup\{L(f,P) | P\space is \space a \space partition\space of \space [a,b]\}$

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My Question:

In our solution our professor has $L(f,P) > 0$.

But if $L(f)$ is $0$ and is also the supremum of $L(f,P)$ then does that not make 0 an upper bound of $L(f,P)$ and thus make it impossible for $L(f,P)$ to be greater than $0$?

In my mind I could simple show that $L(f,P) \le 0$ and since $f(x) \ge 0$ then $f(x) = 0$ and $L(f,P) = 0$.

Thank you in advance. At present I believe it may be a typo. But I shared my logic above and if there is an error in my thinking please let me know. I greatly appreciate it. :)

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Edit:

I forgot to specify. My professor has this line;

Let $P = \{a, c- \frac{\delta}{2},c + \frac{\delta}{2},b\}$ be a partition of $[a,b]$ Since $f(x) > 0$ for all $x \in [c - \frac{\delta}{2},c+\frac{\delta}{2}]$, we conclude that $L(f,P) > 0$.

Answer 1

I think its a proof by contradiction. You suppose $f \neq 0$ so $\exists (\alpha,\beta) \in [a,b]^2, \forall x \in [\alpha,\beta]< f(x) > 0$ since $\forall x \in [a,b], f(x) \geq 0$

You can suppose there is only one such interval, the reasoning being the same should there be more. So $\forall x \in [a,\alpha)\cap (\beta,b], f(x) =0$

Thus since f is continuous (hence the inf exists and is different from $0$): $$L(f,P) = (\beta - \alpha)\inf\{f(x), x \in [\alpha,\beta]\} > 0$$

However $ 0 = L(f) \geq L(f,P) > 0$ There is contradictions, so $\forall x \in [a,b], f(x) = 0$