# An Incorrect Construction of the Ito Integral

Let $$B_t$$ be a Brownian motion defined on the interval $$[0,T]$$, with underlying (filtered) probability space $$(\Omega,\mathcal{F},\{\mathcal{F}_t\},\mathbb{P})$$. Call a function $$f:[0,T]\times\Omega\to\mathbb{R}$$ elementary if it is of the form:

$$f(t,\omega) = X(\omega)\mathbf{1}_{(a,b]}(t)$$

where $$a$$ and $$b$$ are reals and $$X$$ is an $$\mathcal{F}_a$$-measurable random variable. Call a function simple if it is a finite linear combination of elementary functions. Now define the following $$\sigma$$-algebra:

$$\mathcal{P} = \sigma(\{f^{-1}((-\infty,\alpha)) : f \text{ simple}, \alpha \in \mathbb{R} \}),$$

which is a sub-$$\sigma$$-algebra of $$\mathcal{B}([0,T])\times\mathcal{F}$$. Call a function predictable if it is $$\mathcal{P}$$-measurable. Note that $$([0,T]\times\Omega,\mathcal{P},\lambda\times\mathbb{P})$$ can be viewed as a measure space in its own regard. Hence, it makes sense to talk about the Hilbert space $$L^2([0,T]\times\Omega,\mathcal{P},\lambda\times\mathbb{P})$$. However, we are in the setting of the following general result:

Lemma. Let $$(X,\mathcal{A},\mu)$$ be a $$\sigma$$-finite measure space. If $$\mathcal{G} \subseteq \mathcal{A}$$ is a sub-$$\sigma$$-algebra, then $$L^2(X,\mathcal{G},\mu)$$ is a closed linear subspace of $$L^2(X,\mathcal{A},\mu)$$.

From this we construct the Ito integral as follows: Take $$\phi \in L^2([0,T]\times\Omega,\mathcal{P},\lambda\times\mathbb{P})$$ a square-integrable, predictable function. Then let $$f_1,f_2,\dots$$ be a sequence of simple functions with $$f_n\to\phi$$ in $$L^2([0,T]\times\Omega)$$ as $$n\to\infty$$. Using this we define

$$\int_{0}^{T}\phi(s,\omega)dB_s := \lim_{n\to\infty}\int_{0}^{T}f_n(s,\omega)dB_s$$

where the limit on the right is taken in $$L^2(\Omega)$$.

I would like to identify out what is wrong in this construction. If this were valid, then the set of Ito-integrable functions would be a Hilbert space, which is not a statement I have been able to find anywhere....