Proving operator induced by $*$random element is continuous 
Let $X$ be a Banach space and $ 0\leq p\leq \infty$. The operator
  $T\eta :X \to L^p$ induced by a $\ast-$element $\eta: \Omega \to X^*$
  of weak order $p$ is continuous. $T\eta x= x (\eta(\omega))$ where $X$
  is identified with its image in $X^{**}$.

My idea is: $X$ and $L^p$ are both complete metric spaces, so we can use closed graph theorem, but how do I show  ${(x,T_\eta x)}$ is closed in $X\times L^p$.
For more context, $\eta$ having weak order $p$ means
$$\mathbb{E }|\langle x,\eta\rangle |^p<\infty,\quad \forall x\in X.$$
 A: Set $A_N = \{\omega\in\Omega : \|\eta(\omega)\|\le N\}$. Now, let $(x_n)\subset X$ be a sequence and $x\in X$, $f\in L^p$, such that $x_n\to x$ in $X$ and $T_\eta x_n\to f$ in $L^p$. Then, if $1\le p<\infty$,
\begin{align}
&\left(\int_{A_N}|\eta(\omega)x-f(\omega)|^p\,dP\right)^{1/p}\\
&\le\left(\int_{A_N}|\eta(\omega)(x-x_n)|^p\,dP\right)^{1/p} + \left(\int_{A_N}|\eta(\omega)x_n-f(\omega)|^p\,dP\right)^{1/p}\\
&\le\left(\int_{A_N}\|\eta(\omega)\|^p\|x-x_n\|^p\,dP\right)^{1/p} + \left(\int_{\Omega}|\eta(\omega)x_n-f(\omega)|^p\,dP\right)^{1/p}\\
&\le N\|x-x_n\| + \|T_\eta x_n-f\|.
\end{align}
Hence, $T_\eta x = f$ a.e. on $A_N$ for each $N$ and so $T_\eta x = f$ on $\Omega = \bigcup_{N\in\mathbb N}A_N$. By the closed graph theorem it follows that $T_\eta$ is a bounded operator.
If $0<p<1$, the above estimate holds with the exponent $1/p$ omitted and each summand in the last line taken to the $p$. The result thus is the same. For $p=\infty$ you can use the same argument as above. Since I don't know much about $L^0$ (and I don't want to right now) I leave this case to you.
