# Levy's extension of the Borel-Cantelli Lemmas

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 $Z_n / Y_n \rightarrow 1$.

Look at section 12.14(a) on the same page 124,

"We see that $$\langle W\rangle_n \leq 1$$ , a.s., so that lim $$W_n$$ exists, a.s.. Kronecker's Lemma now shows that (a) $${{M_n}\over{A_n}} \to 0$$ a.s. on { $$A_\infty = \infty$$ }."

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 {$$A_\infty = \infty$$} .