Given a filtered probability space $(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\geq0},P)$. Let $1\leq p < \infty, $ a process $x(t)$ $\in \mathcal{L}^p([0,T],R^d)$ $\Leftrightarrow$ $\{ x(t), t \in [0,T]\}$ is $\mathcal{B}([0,T])\times\Omega \rightarrow \mathcal{B}(R^d)$ measurable, $\mathcal{F}_t$-adapted, $R^d$-valued processe such that:
$$\int_{0}^{T} |x(t)|^p dt < \infty, a.s.$$
Let $x(t) \in \mathcal{M}^p([0,T],R^d)$ $\Leftrightarrow$ $x(t) \in \mathcal{L}^p([0,T],R^d)$ and satisfies: $$E\int_{0}^{T} |x(t)|^p dt < \infty.$$
It is easy to check that $\mathcal{M}^p([0,T],R^d)$ is a normed vector space with the norm: $$||x||_p = \left(E\int_{0}^{T}|x(t)|^p dt \right)^{1/p} < \infty.$$ Here is the question:
Why $\mathcal{M}^p([0,T],R^d)$ is a Banach space?
My idea:
Let $\{x_n(t)\}_{n\geq 1} \subset \mathcal{M}^p([0,T],R^d)$ a Cauchy sequence, then we know that there exists an element $x(t)$ in $L^p([0,T]\times \Omega, \mathcal{B}([0,T])\times\Omega, dt\times dP)$(normal product measure space) such that: $$E\int_{0}^{T}|x_n(t) - x(t)|^p dt \rightarrow 0,\ \ \ \ (n \rightarrow \infty).$$
We want to show $\mathcal{M}^p([0,T],R^d)$ is a Banach space, then we need to prove that $x(t)$ must be in $\mathcal{M}^p([0,T],R^d)$. How do we know $x(t)$ is $\mathcal{F}_t$-adapted process from the condition $E\int_{0}^{T}|x_n(t) - x(t)|^p dt \rightarrow 0,(n \rightarrow \infty)$.