if f > g on T, then so are the integrals?

this has me confused right now. anyone have any tips on how to prove in general that

if f(x) > g(x) \forall x \in T, then

\int_T f\, dx > \int_T g\, dx

it seems obvious that it's true when you plot it and look at it and there's where my trouble is. how can i get past it being "obvious" and prove it?

What we need to find is a function that's strictly positive that is Riemann-integrable with integral 0. Basically, it'd be something similar to this.

If for rational x = p/q ( in simplest form) you let f( x)  = 1/q, and f( x) = 0 for x irrational, then it can be shown that that f is integrable with value 0. Now what needs to be done is change the irrational values of f so that they are positive, but have a similar pattern to how we defined f at the rationals. Something like that would be a counterexample, though I can't think of a simple way to represent it explicitly. Maybe one could define it based on the decimal expansions.

I'm now pretty sure that what the OP is trying to show is incorrect.