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....


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