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

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?
