I'm working on Exercise 10.4.10 in Donald Cohn's book Measure Theory. The Exercise is the following:
Let $(\Omega,\mathscr{A},P)$ be a probability space, let $\{X_n\}$ be a sequence of independent random variables, let $S_N = \sum_{1}^{N} X_n$ be the corresponding partial sums, with distribution $P_{S_N} = PS_N^{-1} =:\mu_N$ and finally let $\{\phi_{\mu_N}\}$ be the sequence of characteristic functions $\phi_{\mu_N}(t) = \int_{\mathbb{R}}e^{ixt}d\mu_N(x)$. Prove that
- If $\lim_n\phi_{\mu_N}(t)$ exists and is non-zero for a set of positive measure
- Then $S_N$ converges almost surely.
We are given two Lemmas:
Lemma 1: If $\{a_n\}$ is a sequence of real numbers such that $e^{ita_n}$ converges for all $t$ in a set of positive Lebesgue measure then $a_n$ is convergent.
Lemma 2: If for a fixed $t$ the limit $\lim E(e^{itS_n})$ exists and is non-zero then $\{e^{itS_n}\}$ converges almost surely.
Since $\phi_{\mu_N}(t) = E(e^{itS_N})$ we are in the situation of applying Lemma 2. This Lemma says that $\lim_N e^{itS_N}$ exists almost surely on a set of $t$'s with positive Lebesgue measure, denote this set with $T$. Now I would like to apply Lemma 1, however this is not immediately possible. For this to work, we need to know that there is a set $A\subseteq \Omega$ with $P(A) = 1$ such that for any $\omega \in A$ the limit $\lim_N e^{itS_N(\omega)}$ exists for all $t\in T$. I don't see how this conclusion can be drawn.
By the basic hypothesis there is a set $H\subset\Bbb R$ with $0<\lambda(H)<\infty$ ($\lambda$ is Lebesgue measure) such that $\lim_nE[e^{itS_n}]$ exists and is positive for each $t\in H$. By Lemma 2, if $t\in H$ then there is an event $N_t$ with $P[N_t]=0$ such that $\lim_n e^{itS_n(\omega)}$ exists for all $\omega\notin N_t$.
Consider the (product measurable) set $G:=\{(\omega,t): \lim_n e^{itS_n(\omega)}$ exists$\}$. Observe that for each $t\in H$, $N_t^c$ (the complement of $N_t$) is a subset of the section $$ G^{(t)}:=\{\omega: (\omega,t)\in G\}. $$ It follows that $P[G^{(t)}]=1$ for each $t\in H$. Consequently, by Fubini's theorem, $$ \eqalign{ P\otimes\lambda[(\Omega\times H)\cap G] &=\int_{H}\left[\int_\Omega 1_G(\omega,t)\,P(d\omega)\right]\,\lambda(dt)\cr &= \int_{H}P[G^{(t)}]\,\lambda(dt)\cr &=\lambda(H)=P\otimes\lambda[\Omega\times H].\cr } $$ It follows that $P\otimes\lambda[(\Omega\times H)\setminus G]=0$.
Define $K:=\{\omega: \lambda(G_{(\omega)})>0\}$. We know from Lemma 1 that if $\omega\in K$ then $\lim_n S_n(\omega)$ exists. To conclude we need to check that $P[K]=1$; or, equivalently, that $P[K^c]=0$. Using Fubini again, $$ \eqalign{ 0 &=P\otimes\lambda[(\Omega\times H)\setminus G]\cr &=\int_\Omega\lambda(H\setminus G_{(\omega)})\,P(d\omega)\cr &\ge\int_{K^c}\lambda(H\setminus G_{(\omega)})\,P(d\omega)\cr &=\int_{K^c}\lambda(H)\,P(d\omega) = \lambda(H)P(K^c).\cr } $$ (The next-to-last equality follows from the fact that $\lambda(G_{(\omega)})=0$ if $\omega\in K^c$.) Because $\lambda(H)>0$, we must have $P[K^c]=0$, as desired.