Extension of Riemann-Lebesgue Lemma I'm working my way through Billingsley's Probability and Measure, and I've got a question on Problem 26.5: Show by Theorem 26.1 (Riemann-Lebesgue Theorem) and integration by parts that if $\mu$ has a density $f$ with integrable derivative $f'$ then $\phi(t)=o(t^{-1})$ as $|t|\rightarrow\infty$.  Here $\phi(t)$ is the characteristic function.  
Integration by parts gives me $$\phi(t)=\int_{-\infty}^\infty f(x)e^{itx}dx=\left[f(x)e^{itx}/it\right]^\infty_{-\infty}-\int_{-\infty}^\infty f'(x)e^{itx}/it dx$$
Multiplying both sides by $t$ then gives $t\phi(t)=-\left[f(x)e^{itx}i\right]^\infty_{-\infty}+\int_{-\infty}^\infty f'(x)e^{itx}i dx$$
We want to show the right hand side goes to 0 as $t\rightarrow\infty$.  The second term goes to 0 as $|t|\rightarrow\infty$ since $f'$ is integrable (for this we can use the same argument used to prove the Riemann-Lebesgue Lemma).  But how about the first term?  Do we know that $f(x)\rightarrow 0$ as $|x|$ goes to $\infty$?  I know that generally for $f(x)$ to be integrable this is not required.  Is it required if $f'$ is integrable?  If so, can somebody please give a hint as to why this is so?  
Thanks!

According to some form of the fundamental theorem of calculus (Theorem 7.21 of Rudin's Real and Complex analysis), if $f$ is differentiable and its derivative is integrable, then $f(x) = f(0) + \int_0^x f'(t)dt$ , which converges (by dominated convergence theorem for instance) to $f(0)+ \int_0^{\pm\infty} f'(t)dt \in \mathbb R$ when $x$ goes to $\pm\infty$. These two limits must be $0$ otherwise $f$ couldn't be a density.
