I am having trouble with a particular step, in the Oksendal book (Intro to SDE), exercise 4.10.
It is said to apply Ito isometry to $$\int_0^t\frac{W_s}{\epsilon}\mathbb{1}_{W_s \in \left(-\epsilon, \epsilon\right)}\, dW_s$$ For $\epsilon \in (0,1)$, and then let $\epsilon \rightarrow 0$ to see that the integral vanishes.
However applying the isometry $$\mathbb{E}\left[\left(\int_0^t\frac{W_s}{\epsilon}\mathbb{1}_{W_s \in \left(-\epsilon, \epsilon\right)} \, dW_s\right)^2\right]=\mathbb{E}\left[\int_0^t\left(\frac{W_s}{\epsilon}\mathbb{1}_{W_s \in \left(-\epsilon, \epsilon\right)}\right)^2 \, ds\right]=\frac{\int_0^t\mathbb{E}\left[W_s\mathbb{1}_{W_s \in \left(-\epsilon, \epsilon\right)}\right]^2 \, ds}{\epsilon^2}$$ And let's say at this point that this were correct, I hardly see how this could not blow up as $\epsilon \rightarrow 0$.... A nudge in the good direction would be appreciated.
Since
$$\mathbb{E}(W_s^2 1_{\{W_s \in (-\epsilon,\epsilon)\}}) \leq \epsilon^2 \mathbb{E}(1_{\{W_s \in (-\epsilon,\epsilon)\}}) = \epsilon^2 \mathbb{P}(-\epsilon < W_s < \epsilon)$$
we have
$$\frac{1}{\epsilon^2} \int_0^t \mathbb{E}([W_s 1_{\{W_s \in (-\epsilon,\epsilon)\}}]^2) \, ds\leq \int_0^t \mathbb{P}(-\epsilon<W_s < \epsilon) \, ds. \tag{1}$$
As
$$\mathbb{P}(-\epsilon < W_s < \epsilon) \xrightarrow[]{\epsilon \to 0} \mathbb{P}(W_s = 0)= 0$$
for any $s>0$, an application of the dominated convergence theorem shows that the right-hand side of $(1)$ converges to $0$ as $\epsilon \to 0$. This finishes the proof.