If the difference of the function only dependent on the difference of input, can we say it's linear without assuming it's continuous? As state in the title, if for a function $$f(t+\tau)-f(t)=R(\tau), \forall t,\tau$$
$R(\tau)$ is just a general function showing the difference has nothing to do with $t$.
then I can prove $f(t)=at+b$ where $a,b$ is some certain constant, if assuming $f$ continuous, I prove it by first proving it's true over rational number set.
However, is this also true even if we don't specify the function to be continuous but only measurable? 

Yes, measurability implies that $f$ is affine. Proof: Let $g(t) = f(t) - f(0)$. Then $g$ is additive, because $$ g(a + b) = f(a + b) - f(0) = f(a) + R(b) - f(0),$$ and $R(b) = f(0 + b) - f(0)$, so $g(a + b) = f(a) + f(b) - 2f(0) = g(a) + g(b)$. Furthermore, $g$ is measurable. Every additive, measurable function is linear (see e.g. Theorem 5.5 in Horst Herrlich's "Axiom of Choice".) Therefore, there is some $a$ so that $g(t) = at$ for all $t$, and hence $f(t) = at + f(0)$ for all $t$.
