Ito's Integral's definition: Importance of isometry I'm reading Oksendal's Stochastic Differential Equations (5th edition). He defines the Ito integral of $f$ as the limit
$$\lim_{n\to\infty} \int^T_S \phi_n(t,\omega) dB_t(\omega)$$
Where $\{\phi_n\}$ are elementary functions that satisfy
$$E\bigg[\int^T_S (f-\phi_n)^2 dt\bigg]\to 0\;\;\;\;\text{ for }n\to\infty$$
(I left out a lot of details for brevity, if they are needed, I will add them upon request)
He then says that thanks to the Ito's isometry
$$E\Big[\Big(\int^T_S \phi_t(\omega)\; dB_t(\omega)\Big)^2\Big]=E\Big[\int^T_S
\phi_t(\omega)^2\; dt \Big]$$
the limit in the definition exists and does not depend on $\phi_n$ as long as they satisfy the said condition. 
I don't understand why is the isometry needed. To me it seems like from $\|f-\phi\|_{L^2}\to 0$ and the reverse triangle inequality we already get $\|\phi\|_{L^2}\to\|f\|_{L^2}$ which (for $\sigma$-finite measure spaces) implies convergece in $L^1$. Where am I wrong?
 A: Why are you trying to show convergence in $L^1$ of the $\phi_n$ when what is required is to show $L^2$-convergence of the stochastic integrals $\int^T_S \phi_n \text dB_t$? 
In any case, in order to show $\ \int^T_S \phi_n \text dB_t\ {}\stackrel{L^2}{\to}\int_{S}^{T}f\text dB_{t}\ $ (assuming $\ \mathbb E[\int_{S}^{T}f^2\text d{t}] < \infty$), Oksendal argues that elementary functions $\phi_n$ which approximate $f$ both exist and satisfy $\mathbb E[\int_{S}^{T}\phi_n^2\text d{t}]<\infty$. So, to show $L^2$-convergence, we may proceed by showing that the sequence of stochastic integrals $\int_{S}^{T}\phi_n\text dB_t$ form a Cauchy-sequence in the $L^2$-space of random variables with zero mean, finite second moment and convergence in mean-square. Because this $L^2$-space is complete, this implies that the integrals converge to a random variable that belongs to the space. It is this limit that is defined to be $\int_{S}^{T}f\text dB_{t}$. We proceed by observing:
$\mathbb E[(\int_{S}^{T}\phi_n\text dB_{t})^2] \stackrel{\text{Ito's Iso.}}{=} \mathbb E[\int_{S}^{T}\phi_n^2\text d{t}] < \infty\tag{1}$
And, for $\phi_m$ and $\phi_n$,
$\mathbb E[(\int^T_S \phi_n \text dB_t-\int_{S}^{T}\phi_m\text dB_{t})^2] \ =\  \mathbb E[(\int^T_S (\phi_n - \phi_m)\,\text dB_{t})^2] \\\hspace{5.1cm}\ \stackrel{\text{Ito's Iso.}}{=}\ \mathbb E[\int^T_S (\phi_n - \phi_m)^2\,\text dt]\hspace{5.7cm} \leqslant\ 2\mathbb E[\int^T_S (\phi_n - f)^2\,\text dt] + 2\mathbb E[\int^T_S (\phi_m - f)^2\,\text dt],$
from which the property $\ \lim\limits_{n\to\infty}\mathbb E\left[\int^T_S (\phi_n - f)^2\,\text dt\right]=0\ $ immediately implies 

$$\lim\limits_{n\to\infty}\lim\limits_{m\to\infty}\mathbb E\left[\left(\int^T_S \phi_n \text dB_t-\int_{S}^{T}\phi_m\text dB_{t}\right)^2\right] = 0\,.\tag{2}$$

Together, (1) and (2) imply that the $L^2$-limit of the elementary stochastic integrals, which we choose to define $\int_{S}^{T}f\text dB_t$, exists. As to whether one must use Ito's Isometry to show this, I do not know. But, you have to admit, its use does simplify the argument quite nicely, no?
