Let $g(x) \ge 0$ Riemann integrable on $[a,b]$. Show using Riemann sums that for each subinterval $[c,d] \subset [a,b]$:
$$\int^b_a g(x)dx \ge \int^d_c g(x)dx$$
I thought that we should show that every lower sum of $\int^d_c g(x)dx$ is $\le$ some lower sum of $\int^b_a g(x)dx$. But I don't know how to show it.
Easier still, $$ \int_a^b g(x)\,\mathrm{d}x = \int_a^c g(x)\,\mathrm{d}x + \int_c^d g(x)\,\mathrm{d}x + \int_d^b g(x)\,\mathrm{d}x, $$ so all you need is for the first and third terms of the righthand side to be $\ge 0$. But this is clear: if $f\ge 0$ and integrable on an interval $[u,v]$ then $\int_u^v f(x)\,\mathrm{d}x \ge 0$, because every term of every Riemann sum is $\ge 0$.