$f \geq 0 \ on \ [a,b] \ RI.$
Then, $$\int_{a}^{b} f(x) dx \geq 0$$
I have seen this
Is the Riemann integral of a strictly positive function positive?
However, I'm not fully satisfied with the answers there as the conditions are slightly different. In the accepted answer, they say that since $f$ is continuous and positive on I (to make it similar to my problem let $I = [a,b]$, then $\int_{I} f > 0$. However, this is the conclusion. To me, it seems like they used the final answer to prove the result.
In my problem, however, I am not told $f$ is continuous in my problem. So that is a major assumption I am missing because while every continuous function is RI, being RI does not guarantee continuity.
I did look at a second answer on that same problem that has more upvotes. However, they prove it by assuming that $\int_{0}^{1} f = 0$. This works for that problem because $f > 0$ strictly. In my problem, however, it could be equal to $0$. So showing this case won't really help me here.
There was one more answer, but that uses Lebesgue integration, which I have not covered in this class.
I am really not so sure how to start it myself. I am convinced it's true after trying to come up with counterexamples because I wasn't convinced at first, but I really have no idea how to show it. Can anyone help me with this one? Thanks!
The definition of the Riemann integral looks something like
$$ \sup_{\mathcal{P}} \sum_{i=1}^n \left(\inf_{[x_{i},x_{i+1}]} f(x) \right) (x_i - x_{i-1}) = \int_a^b f = \inf_{\mathcal{P}} \sum_{i=1}^n \left(\sup_{[x_{i},x_{i+1}]} f(x) \right) (x_i - x_{i-1}). $$
You might write this somewhat differently but the point is that every piece of this is $\ge 0$ by assumption. Specifically, if $f \ge 0$ then $\inf f \ge 0$ on any interval.