Norm of a Simple Random Variable

Question

Let $(\omega, \Sigma, P)$ be a measure space and $(B, \|\cdot\|_{B})$ be a Banach space. Assume that $\eta : \Omega \rightarrow B$ is a simple variable defined as, \begin{equation} \eta (\omega)=\sum_{k=1}^{n} c_{k}I_{E_k}(\omega), \; \omega \in \Omega \end{equation} Where $c_k \in B$ and $E_1, E_2, \dots, E_n$ are disjoint random events. Then how the following expression is possible? \begin{equation} \|\eta(\omega)\|_{B}= \sum_{k=1}^{n}\|c_k\|_{B}I_{E_k}(\omega) \end{equation} In my view, if I take the norm on both sides of $\eta(\omega)$, then I should use the triangle inequality property. How this equality holds then?

Answer 1

Any $\omega$ can belong to at most one $E_k$ by disjointness. So there is actually only one term in the sum for any particualr $\omega$. Do you now see why equality holds?

Let $\omega \in E_i$. Then $\eta (\omega)=c_i$ since only one term ( the one corresponding to $k=i$) in the sum defining $\eta (\omega)$ is non-zero. Hence, $\|\eta (\omega)\|_B=\|c_i\|_B$. Also, $\sum_{k=1}^{n}\|c_k\|_{B}I_{E_k}(\omega)=\|c_i\|_B$ since, once gain, only one term is non-zero.