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?

Suppose f and g are integrable and f > g for all x in T.

Then f = g + h where h > 0
h = f-g
Integral of h over T is INT F - INT G using linearity of integrals.
Thus, INT H = INT F - INT G
INT F = INT G + INT H

Since H > 0 for all x, any lower sum be positive, and this is less than the value of the integral. Thus INT H > 0 and INT F > INT G.