How to modify the conditions so that the theorem is true. 
Let $f(x)$ and $g(x)$ be two increasing and differentiable functions from $\Bbb{R}$ to $\Bbb{R}$. If $f'(x)>g'(x)$ $\forall x$ then there exists an interval $[a,\infty)$ for some real number $a$ for which $f(x)>g(x)$.

This theorem is false, as shown by many counter examples. But how do I modify the conditions so that the theorem becomes true.?
 A: No, this isn't true. $g$ could be a line, for instance, and $f$ could increase to it as an oblique asymptote. You need $f'>g'+\epsilon$ for some positive $\epsilon$, in which case the fundamental theorem of calculus will get you the result right away.
A: You could reformulate your proposition as follows, by considering $h = f - g$:  If $h$ is a differentiable real function such that $h'(x) > 0$ for all $x$ then necessarily there exists a $c$ such that $h(c) > 0$.
Note that if $h(c)>0$, since $h' >0$ it will be true that $h(x) > h(c) >0$ for all $x > c$, so that element of your proposition has been captured as well.
When you write it this way I think you can see the improbability of this claim.  Does a counterexample come to mind?
Although it is not the least assumptions, this proposition would be correct if the function is assumed to be both increasing and concave up - since a concave up function sits above its tangent lines, and the tangent lines have positive slope, the function must eventually be positive.
e: I'd like to include Kevin's condition as well - if $h'(x)$ is not merely positive but $h'(x) > \epsilon$ for some $\epsilon > 0$, then the proposition holds as well.
A: Let $h(x)=f(x)-g(x)$ .
$h'(x)>0$ so $h(x)$ is strictly increasing.
Let $x>a$ . We know that $$h(x)>h(a)\Rightarrow f(x)-g(x)>f(a)-g(a)\Rightarrow f(x)>g(x)+f(a)-g(a)$$ 
If $f(a)-g(a)\geq 0$
 we could conclude that $ f(x)>g(x)$ for every $x$ greater than $a$.
So, a sufficient condition is that there is an $a$ for which $f(a)\geq g(a)$ holds.
