Prove that $\underline\int_{a}^bf\ge0$ when $f(x)\ge 0$

Question

My question is: Suppose that the bounded function $f:[a,b]\rightarrow \Bbb R$ has the property $f(x)\ge 0$ for all $x$ in $[a,b]$. Prove that $\underline\int_{a}^bf\ge0$. I am just confused on where to start. I would think you have to use the definition of a lower integral but I am unsure how to apply it if it is even the right step forward.

Answer 1

Hint:

We know that the lower integral is defined as the supremum of the lower sums over all partitions.

So, it suffices to show that there exists a partition whose lower sum is non-negative, as the supremum of lower sums is at least as large as any particular lower sum.

So, can you pick ANY partition, (say, $\{a,b\}$, as @yanko suggested in the comments), and show that the lower sum on that partition must be non-negative?