Why is the isometry of It\^o integral called so? For functions $f$ satisfying appropriate (good) conditions, the following property is called to be isometric.

$$
E\left[\left|\int_{a}^{b}f(t,\omega)dB_{t}(\omega)\right|^{2}\right]=E\left[\int_{a}^{b}|f(t,\omega)|^{2}dt\right]
$$

Here $B_{t}(\omega)$ is the Brownian motion.
Why is this called to be isometric?
If one defines the $L^{2}$-norm by
$$
\|f\|_{L^{2}(P)}=(E[|f|^{2}])^{1/2},
$$
then the above equality is rewritten as
$$
\left\|\int_{a}^{b}f_{t} dB_{t}\right\|_{L^{2}(P)}^{2}=\int_{a}^{b}\|f\|_{L^{2}(P)}^{2}(t)dt
$$
provided that Fubini theorem is valid.
However I don't look isometry as in functional analysis in Hilbert space, i.e., $\|Tf\|_{H_{1}}=\|f\|_{H_{2}}$ for some operator $T$.
What I don't connect between Ito integral and Riemann integral is also a reason that I don't know why it's called to be isometric.
I'm glad if you tell me a reason.
 A: Consider a function $f: [0,T] \times \Omega \to \mathbb{R}$ which is also in your "good conditions" space, denoted $\mathcal{V}$. Consider the operator $I : \mathcal{V} \to L^2(\Omega, \mathcal{F}_T, P)$ given by
$$
I_T(f) := \int_0^T f(s,\omega) \, dB_t(\omega).
$$
Ito's isometry tells us
$$
E\left( \left( \int_0^T f(s) \, dB_t \right)^2 \right) = \int_0^T E(f^2(s)) ds.
$$
Now, note
\begin{align}
||f||_{L^2([0,T] \times \Omega)}^2 & = \int_{[0,T] \times \Omega} f^2(s,\omega) \, d(Leb \otimes P)(s,\omega) \\
& = \int_0^T \int_{\Omega} f^2(s,\omega) \,dP(\omega) \, ds\\
& = \int_0^T E (f^2(s)) ds \\
& = \int_{\Omega} \left( \int_0^T f(s,\omega)\, dB_t(\omega)\right)^2 \, dP(\omega) \\
& = \int_{\Omega} \left(I_T(f)(\omega)\right)^2 \, dP(\omega) \\
& = ||I_T(f)||_{L^2(\Omega)}^2.
\end{align}
The second equality is Tonelli's theorem, valid since $f^2 \geq 0$ and both measures are $\sigma$-finite and the fourth equality is Ito's isometry.
Writing out the entire measure spaces, the opreator $I_T$ is an isometry from
$$
L^2([0,T] \times \Omega, \mathcal{B} \otimes \mathcal{F}_T, Leb \otimes P) \to L^2(\Omega, \mathcal{F}_T, P),
$$
so long as your function $f \in \mathcal{V}$.
