Following is the statement and proof of Levy's extension of the Borel-Cantelli Lemmas, as given in Williams' "Probability with Martingales" (1991), in section 12.15 on page 124. I understand most of the proof, except for the part "then $M_n / A_n \rightarrow 0$" near the end, which I have emboldened (it was not bold in the original). Please help me to see why this statement holds. Thank you.
Assume $\left(\mathcal{F}_n\right)$ is a filtration.
Theorem
Suppose that for $n \in \mathbb{N}$, $E_n \in \mathcal{F}_n$. Define
$$ Z_n := \sum_{1 \leq k \leq n}\mathbb{1}_{E_k} = \textrm{number of } E_k\ \left(k \leq n\right)\ \textrm{which occur.} $$Define $\xi_k := P\left(E_k \mid \mathcal{F}_{k – 1} \right)$, and
$$ Y_n := \sum_{1 \leq k \leq n}\xi_k $$Then, almost surely,
a. $\left(Y_\infty\right) \implies \left(Z_\infty < \infty\right)$,
b. $\left(Y_\infty = \infty\right) \implies \left(Z_n / Y_n \rightarrow 1\right)$.
Proof
Let $M$ be the martingale $Z – Y$, so that $Z = M + Y$ is the
decomposition of the submartingale $Z$. Then (you check!) $$ A_n := \left<M\right>_n = \sum_{k \leq n}\xi_k\left(1 – \xi_k\right) \leq Y_n\ \textrm{a.s.} $$If $Y_\infty < \infty$, then $A_\infty < \infty$ and $\lim M_n$
exists, so that $Z_\infty$ is finite. (We are skipping 'except for a
null $\omega$-set' statements now.)If $Y_\infty = \infty$ and $A_\infty < \infty$ then $\lim M_n$ exists
and it is trivial that $Z_n / Y_n \rightarrow 1$.If $Y_\infty = \infty$ and $A_\infty = \infty$, then $\mathbf{M_n / A_n \rightarrow 0}$, so that, a fortiriori, $M_n / Y_n \rightarrow 0$ and $Z_n / Y_n \rightarrow 1$.
Best Answer
Look at section 12.14(a) on the same page 124,
I think you want to know why ${{M_n} \over {A_n}} \to 0 $ by Kronecker's Lemma. The deduction is as follows .
Since $$ W_n = \sum_{k=0}^n {a_k\over b_k} = \sum_{k=0}^n {{M_k - M_{k-1}}\over {1 + A_k}}, $$ so $$ a_n = M_k - M_{k-1} , b_n = 1 + A_k, $$ now $$ S_n = \sum_{k=0}^n a_k = M_n ,$$ Kronecker's Lemma follows to show $$ {S_n \over b_n} = {M_n \over { 1 + A_n }} \to 0, $$, which is equivalent to state $ {M_n \over A_n} \to 0 $ on {$ A_\infty = \infty $} .