I am trying to understand Durrett's proof of the continuity theorem. It contains two parts. The one I have problems with is the second part, whose statement is: If $\varphi_n(t)$ converges pointwise to a limit $\varphi(t)$ that is continuous at $0$, then the associated sequence of distributions $\mu_n$ is tight and converges weakly to the measure $\mu$ with characteristic functions $\phi$. Here the $\mu_n$'s are probability measures and the $\varphi_n$'s are the associated ch.f.
I have two minor questions about Durrett's proof; both are highlighted in a darker blue:
(i) In the proof of tightness, Durret picked a $u$ such that $\int_{-u}^u(1-\varphi(t))dt < \epsilon$. But then he claimed $2\epsilon \geq u^{-1}\int_{-u}^u(1-\varphi_n(t))dt$ for $n \geq N$ since $\varphi_n(t) \rightarrow \varphi(t)$ for each t.
I understand the convergence part, but where does this $2\epsilon$ come from? Why does $2 \geq u^{-1}$?
(ii) The second part is that he claimed that the "last observation" (which I suppose is the previous sentence that if $\mu_{n(k)} \Rightarrow \mu$ then $\mu$ has a ch.f. $\varphi$) and tightness imply that every subsequence has a further subsequence that converges to $\mu$.
I understand that the tightness condition implies every subsequential limit is the distribution function of a probability measure, but why does these two observations imply that every subsequence has a further subsequence that converges to $\mu$?
Thank you very much in advance.
Here are the screenshots of the theorem and the proof.
Best Answer
For (i), you may have misread this part since we have $u^{-1}\int_{-u}^u(1-\varphi(t))\,dt \to 0$ as $u\to 0$ by the continuity of $\varphi$ at $0$, (note the factor of $u^{-1}$ in front). To see the $2\epsilon$, pick $u$ so small that $u^{-1}\int_{-u}^u(1-\varphi(t))\,dt < \epsilon$. By bounded convergence, $u^{-1}\int_{-u}^u(1-\varphi_n(t))\,dt \to u^{-1}\int_{-u}^u(1-\varphi(t))\,dt$, so if $N$ is chosen large enough, we guarantee $|u^{-1}\int_{-u}^u(1-\varphi_n(t))\,dt-u^{-1}\int_{-u}^u(1-\varphi(t))\,dt|<\epsilon$ for all $n\ge N$, hence the claim follows by the triangle inequality.
For (ii), I remember having this same question when I read Durrett. The Key Point is to show that for every bounded continuous function $f$, every subsequence of $\int f\,d\mu_n$ has a further subsequence converging to $\int f\,d\mu$ (i.e. the weak convergence $\mu_n\Rightarrow \mu$). If this is established, then the rest of the claim follows by Durrett's argument.
We now give the proof. The three ingredients of this proof are Helly's selection theorem (Theorem 3.2.12), Theorem 3.2.13 (consequence of tightness) and the inversion formula (Theorem 3.3.11). By Theorem 3.2.12, every subsequence of $\mu_n$ has a further subsequence $\mu_{a(n)}$ that converges vaguely to some limiting distribution $\mu_{a(\infty)}$. By Theorem 3.2.13, $\mu_{a(\infty)}$ is in fact a probability measure (with ch.f. $\varphi(t)$) because it's been proved $\mu_n$ is tight. Because $\mu_{a(\infty)}$ is a probability measure with ch.f. $\varphi(t)$, the inversion formula (Theorem 3.3.11) implies that $\mu_{a(\infty)} = \mu$. This finishes the proof of the Key Point.