A question on the extension of of integrants from simple processes t0 $L^2$? I have a question. While defining the Stochastic integral w.r.t to the Brownian Motion we begin with simple processes which are adapted and left continuous and then extend it to the square integrable processes $L^2(R_+ \times \Omega,\mathcal{P_r},ds \times\mathbb{P})$ which are progressive .
Let $\mathcal{E}$ refer to the set of all simple processes
Now clearly $\mathcal{E}\subset L^2(R_+ \times \Omega,\mathcal{P_r},ds \times\mathbb{P}) $
and the map $$I:\phi \mapsto \int_ 0^{\infty} \phi_s dB_s $$ is a linear isometry from E (endowed with the norm $\|.\|_{L^2(R_+ \times \Omega,\mathcal{P_r},ds \times\mathbb{P})})$ into $L^2(\Omega,\mathcal{F},\mathbb{P})$
Now the claim is that I can be uniquely extended to a linear isometry from the closure of $\mathcal{E}$ into $L^2(\Omega,\mathcal{F},\mathbb{P})$.
Is it because as $\bar{\mathcal{E}}$ is dense in $L^2(R_+ \times \Omega,\mathcal{P_r},ds \times\mathbb{P})$ and since $\phi$ is a bounded linear operator(I am not so sure about it as i have not studied any functional analysis) we can apply the BLt theorem and say that the operator is closable and uniquely extended to the new class of integrands?
Thank you
 A: 
Lemma: Let $(X,\|\cdot\|_X)$ and $(Y,\|\cdot\|_Y)$ be Banach spaces and $D \subseteq X$ dense. If $f:D \to Y$ is an isometry, i.e. $$\|f(x)-f(y)\|_Y = \|x-y\|_X$$ for all $x,y \in D$, then $f$ admits a unique continuous extension which is again an isometry.

Proof: For any $x \in X$, we can choose $(x_n)_n \subseteq D$ such that $x_n \to x$. Then it follows from the fact that $f$ is an isometry on $D$ that $(f(x_n))_{n \in \mathbb{N}}$ is a Cauchy sequence. Since $Y$ is complete, there exists $$\tilde{f}(x) := \lim_{n \to \infty} f(x_n).$$


*

*$\tilde{f}$ is well-defined (i.e. does not depend on the approximation sequence): Let $(x_n)_n \subseteq D$ and $(\tilde{x}_n)_n \subseteq D$ such that $x_n \to x$ and $\tilde{x}_n \to x$. Using again that $f$ is an isometry on $D$, we get $$\|f(x_n)-f(\tilde{x}_n)\|_Y = \|x_n-\tilde{x}_n\|_X \to 0,$$ hence, $$\lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} \tilde{f}(x_n).$$ In particular, $f(x) = \tilde{f}(x)$ for all $x \in D$.

*$\tilde{f}$ is an isometry: Let $x,y \in X$ and $(x_n)_n \subseteq D$, $(y_n)_n \subseteq D$ such that $x_n \to x$, $y_n \to y$. Then $$\|\tilde{f}(x)-\tilde{f}(y)\|_Y = \lim_{n \to \infty} \|f(x_n)-f(y_n)\|_Y = \lim_{n \to \infty} \|x_n-y_n\|_X = \|x-y\|_X.$$


Now let $g$ be an arbitrary continuous extension of $f$. Then, as $f(x_n)=g(x_n)$ for any $x_n \in D$, it follows easily from the continuity of $g$ and $\tilde{f}$ that $\tilde{f}(x)=g(x)$ for any $x \in X$. This finishes the proof.

Apply the above Lemma for $$\begin{align*} X &:= L^2([0,\infty) \times \Omega, \mathcal{P}, ds \otimes \mathbb{P}) \\ D &:= \mathcal{E} \\ Y &:= L^2(\Omega,\mathbb{P}), \\ f(\phi) &:= \int_0^{\infty} \phi_s \, dB_s. \end{align*}$$
