# Almost sure convergence in stochastic integrals

Consider a mesh $\{t_j^{(n)}\}$ for $1\le j\le k(n)$.

Mörters and Peres (Brownian motion, Remark 7.7) give a condition that turns out to be equivalent to
$$\tag{1}\label{1} \sum_{n=1}^\infty \sum_{j=1}^{k(n)} (t^{(n)}_{j+1} – t^{(n)}_j)^2 <\infty$$
(which is used explicitly in Exercise 1.16) and show that it guarantees that almost surely
$$\sum g(B_{t_j})\Delta B_{t_j} \rightarrow \int g(B_s)dB_s$$
where $\Delta B_{t_j} = B_{t_{j+1}} – B_{t_j}$ and $\Delta t_j = t_{j+1}-t_j$. In other words, this sequence of random variables almost surely converges to its $L^2$ limit.

(This is not guaranteed in general. Consider an independent sequence where $X_n=1$ with probability $1/n$ and $X_n=0$ otherwise. Then $\mathbb E(X^n_2)=1/n^2$ so $X_n$ converges to 0 in $L^2$. But $X_n$ does not converge to 0 almost surely, by the second Borel-Cantelli lemma.)

Instead of giving the details on that*, we will give details on something Mörters and Peres do not give as much detail on but which will also reveal how to do the above: what condition on the mesh is required for almost sure convergence of
$$\sum g(B_{t_j})(\Delta B_{t_j})^2 \rightarrow \int g(B_s)ds,$$
or equivalently, since $\int g(B_s)ds$ is a non-stochastic (but random) integral where the mesh does not matter,
$$b_n:= \sum g(B_{t_j})[(\Delta B_{t_j})^2 - \Delta t_j] \rightarrow 0.$$
They show that such convergence happens in $L^2$ without any assumption on the mesh (as most authors do) but then they are a little vague about the rest in Theorem 7.13. So as an exercise we can go back to Remark 7.7 and see if a similar condition can be obtained in this case.

By second-semester calculus it suffices to show that $\sum b_n^2<\infty$ almost surely:
$$\{\omega: \sum b_n(\omega)^2<\infty\}\subseteq\{\omega: b_n(\omega)\rightarrow 0\}.$$
For this it suffices to show that $\sum \mathbb E (b_n^2)<\infty$.

(Indeed, if $\sum \mathbb E (b_n^2)<\infty$ then [by Monotone Convergence] we have $\sum \mathbb E (b_n^2) = \mathbb E (\sum b_n^2)$ and so $\sum b_n^2$ is a nonnegative random variable with finite mean and hence $\sum b_n^2$ must be finite almost surely.

To prove that a nonnegative random variable with finite mean must be finite almost surely, note that for any random variable $X$, if $\mathbb P(X^2\ge M) \ge N/M$ then
$$\mathbb E(X^2)\ge \mathbb E(X^2 | X^2\ge M) \mathbb P(X^2\ge M) \ge M \mathbb P(X^2\ge M) \ge N.$$
Thus, if $\mathbb E(X^2) < N$, where $N$ is a given constant, then $\mathbb P(X^2\ge M) < N/M$.
Consequently, $\mathbb P(X^2=\infty)=0$.

Let $U_M = \{\omega: \sum b_n(\omega)^2\ge M\}$; then $\mathbb P(U_M) < N/M$. We may note that $U_M$ is computably enumerable only relative to the mesh. And indeed by an exercise in MP, the mesh must be computable in order to work for all ML-random $\omega$. With a nonstandard analysis construction of the Ito integral, however, we can sidestep this requirement.)

Calculating exactly as on the bottom of page 47 in Øksendal (6th edition)’s proof sketch for Ito’s lemma, with his $a_j$ being our $g(B_{t_j})$, we have
$$\mathbb E(b_n^2) = 2\sum_{j=1}^{k(n)} \mathbb E(g(B_{t_j})^2) (\Delta t_j)^2.$$
Here the factors $\mathbb E(g(B_{t_j})^2)$ will be bounded by a constant $c$, assuming we are integrating over a bounded domain $\int_S^T$.

The mesh will depend on $g$ in general. Consider for example $g(b)=e^{b^2/2}$ at time $t=1$; then $\mathbb E(e^{B_1^2})=+\infty$.

So our condition is
$$\sum_{n=1}^\infty 2\sum_{j=1}^{k(n)} \mathbb E(g(B_{t_j})^2) (\Delta t_j)^2 \le \sum_{n=1}^\infty 2\sum_{j=1}^{k(n)} c (\Delta t_j)^2 <\infty$$
which is exactly the same as condition (\ref{1}).

*Actually here are some details. We seek to relate the condition
$$\sum_{n=1}^\infty \mathbb E \int_0^t (H_n(s)-H(s))^2 ds <\infty$$
to the mesh. Here $H(s)=g(B_s)$ and $H_n(s)$ is an elementary approximation. The $n$th term is a sum over $j$ of:
$$\mathbb E \int_{t_j}^{t_{j+1}} \left(g(B_s) – \sum_{j=1}^{k(n)} g(B_{t_j}) \chi_{[t_j,t_{j+1})}(s)\right)^2 ds$$
$$= \mathbb E\int_{t_j}^{t_{j+1}} (g(B_s) - g(B_{t_j}))^2 ds = \int_{t_j}^{t_{j+1}} \mathbb E ((g(B_s) - g(B_{t_j}))^2) ds$$
(by dominated convergence)
$$\le c \int_{t_j}^{t_{j+1}} \mathbb E ((B_s) - B_{t_j})^2) ds$$
(since $\mathbb E[(g(B_s)-g(B_{t_j})^2] =\mathbb E\left[\left(\frac{g(B_s)-g(B_{t_j})}{B_s-B_{t_j}}\right)^2(B_s-B_{t_j})^2\right] \le\mathbb E\left[\left(\frac{g(B_s)-g(B_{t_j})}{B_s-B_{t_j}}\right)^2\right]\mathbb E[(B_s-B_{t_j})^2]$)

$$= c \int_{t_j}^{t_{j+1}} \mathbb E ((B_{s – t_j})^2) ds = c \int_{t_j}^{t_{j+1}} s – t_j ds$$
$$= c \int_0^{\Delta t_j} u du = c (\Delta t_j)^2/2$$
so we get
$$\frac{1}{2} \sum_{n=1}^\infty \sum_{j=1}^{k(n)} (t_j^{(n)}-t_{j+1}^{(n)})^2 < \infty$$
which is the same as condition (\ref{1}).

In the case where each $t_{j+1}^{(n)}-t_{j}^{(n)} = 1/k(n)$, this becomes
$$\sum_{n=1}^\infty 1/k(n) < \infty$$
so we can take $k(n)=n^2$ or, like Paley-Wiener, $k(n)=2^n$.