So I am reading, the Xuerong Mao book about Brownian Motion and Stochastic Integrals, and I came up to this definition
Let $0\leq a < b <\infty$. Denote by $\mathcal{M}^2([a,b], \mathbb{R})$ the space of all real-valued measurable $\{F_t\}$-adapted processes $f=\{f(t)\}_{a\leq t \leq b}$ such that $$ \vert\vert f\vert\vert_{a,b}^2=E\int_a^b \vert f(t)\vert^2dt<\infty.$$ We identify $f$ and $\tilde{f}$ in $\mathcal{M}^2([a,b],\mathbb{R})$ if $\| f-\tilde{f}\|t_{a,b}^2=0$. In this case we say that $f$ and $\tilde{f}$ are equivalent and write $f=\tilde{f}$.
After this, they say that $\vert \vert \cdot \vert \vert_{a,b}^2$ is a metric on $\mathcal{M}([a,b],\mathbb{R})$ and the space is complete. However I think that there is a bit of problems with the notation (and I wanted to verify). It could work if we were working with a norm, however since metrics should be maps $\mathcal{M}^2([a,b], \mathbb{R})\times \mathcal{M}^2([a,b], \mathbb{R}) \to \mathbb{R}$ then, to prove it is a metric we would take $f_1$ and $f_2$ and then define their distance by $\| f_1-f_2\|_{a,b}^2$.
Then, heuristically, all the properties of a metric (symmetric, triangle inequality, and non-negativity) would follow by the properties of the absolute value, monotonicity and linearity of the integral and $E$. In particular, the non-negativity would come from the equivalence given in the definition. Am I wrong, or should I work with it as a norm?
Also, besides the fact that the integrals are finite and so if we have a Cauchy secuence we can change integral with the limit and $\lim_{n\to \infty} f_{1n}(t)-f_{2n}(t)=0$, I haven't come up with other arguments to show that the space is complete. So any formal arguments or corrections will be highly appreciated.