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?



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.

  • $\begingroup$ I don't understand why these limits must be 0. See link for example. $\endgroup$ – gruyb Sep 15 '16 at 9:29
  • $\begingroup$ In your case, the fundamental theorem of calculus implies that there is a limit in $+\infty$ (unlike in the counterexample you mentioned). Now if this limit, which we call $\lambda$, is not 0, then there is no way $f$ could integrate to $1$. Hence $\lambda = 0$. Same reasoning for $-\infty$. $\endgroup$ – justt Sep 15 '16 at 9:34
  • $\begingroup$ You're welcome ! $\endgroup$ – justt Sep 15 '16 at 9:52

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