# the continuity in the proof of Ito integral

This is in regards to constructing the Ito integral, specifically the second step of approximating bounded functions by bounded and continuous functions.

Let $$(\Omega, \mathcal{F}, P)$$ be a probability space and let $$V = V(S,T)$$ be the class of functions $$f: [0,\infty) \times \Omega \to \mathbb{R}$$ such that

1. $$(t,\omega) \mapsto f(t,\omega)$$ is $$\mathcal{B} \otimes \mathcal{F}$$ - measurable, where $$\mathcal{B}$$ is the Borel $$\sigma$$-algebra on $$[0,\infty)$$,
2. $$f$$ is $$\mathcal{F}_t$$-adapted,
3. $$E[ \int_S^T f(t,\omega)^2 dt ] < \infty$$.

Oksendal's 6th ed. of "Stochastic Differential Equations," on page 27, states: enter image description here

Let $$h \in V$$ be bounded. Then there exist bounded functions $$g_n\in V$$ such that $$g(\cdot,\omega)$$ continuous for each $$\omega$$ and $$n$$，and $$\lim_{n \to \infty} E\left[ \int_S^T (h - g_n)^2 dt\right] = 0.$$ $$Proof$$:Let $$|h(t,\omega)|\leq M$$ for all $$(t,\omega)$$.For each $$n$$ let $$\psi_n$$ be a non-negative,continuous function on R such that (i)let $$\psi_n=0$$ for $$x\leq -1/n$$ and $$x\geq 0$$ (ii)$$\int_{-\infty}^\infty\psi_n(x)dx=1$$ Define $$g_n(t,\omega)=\int_0^t \psi_n(s-t)h(s,\omega)ds.$$ Then $$g_n(.,\omega)$$ is continuous for each $$\omega$$ and $$|g_n(t,\omega)|\leq M$$. Since $$h\in \mathcal{V}$$ we can show that $$g_n(t,.)$$ is $$\mathcal{F}_t$$-measurable for all $$t$$.Moreover, $$\int_S^T (h - g_n)^2 dt = 0.~as~n\rightarrow \infty,~for~ each~ \omega$$

In the proof, the author constructs a convolution to make the bounded function h continuous and bounded. I'm not sure why $$g_n(.,w)$$ is continuous for each $$\omega$$,and the most important,why $$\int_S^T (h - g_n)^2 dt = 0.~as~n\rightarrow \infty, for~each~\omega$$

• 1) $g_n$ is a convolution which means that it inherits regularity properties from $\psi_n$. Look up results on continuity of convolutions. 2) The convergence holds for a suitable subsequence $g_{n_k}$ of $g_n$ (recall that $L^2(\mathbb{P})$ convergence implies a.s. convergence of a subsequence). – saz Apr 8 at 9:32
• @saz I'm sorry but for 2), do you mean the $(h-g_n)^2\rightarrow 0$ in $L^2(P)$? Could you be a little more detailed? – user8281063 Apr 8 at 13:22
• Ah, sorry, I got something wrong about 2), forget what I wrote. – saz Apr 8 at 13:34