Where $\mathrm dt$ in $$ \int_{\mathbb{R}}f'(t)\mathrm dt=0 $$ denotes Lebesgue integration.
My first instinct was to use the difference quotient form of the derivative $$ f'(t)=\frac{f(t+h)-f(t)}{h}+o(|h|) $$ but the little o term poses problems in integrating over $\mathbb{R}$.
I am also concerned about using the FTC as $f'$ needn't be continuous, but I think the FTC holds almost everywhere? I also considered potentially proving that $f$ is uniformly continuous, possibly using the fact that the tails of the integral of $f'$ can be made arbitrarily small, thus ruling out the annoying tall narrow triangles function which is integrable and doesn't vanish at $\pm \infty$. However, I am having difficulties working out the details and would appreciate any hints!
edit: All I am given is that $f$ is differentiable and the integrability conditions.
One has to be careful. Consider:
$$ f(x) = \cases{0 & if $x \le 0$ or $x > 1$\cr x & if $0 < x \le 1$\cr}$$ $$ f'(x) = \cases{0 & if $x < 0$ or $x > 1$\cr 1 & if $0 < x < 1$\cr \text{undefined} & if $x = 0$ or $x = 1$\cr} $$ Then $f$ and $f'$ are Lebesgue integrable on $\mathbb R$, but $\int_{\mathbb R} f'(x)\; dx = 1$.
EDIT: On the other hand, it is true that if $f$ is differentiable everywhere and $f' \in L^1[a,b]$, then $$f(b) - f(a) = \int_a^b f'(x)\; dx$$ See e.g. Rudin, "Real and Complex Analysis", Theorem 8.21.