I study Bochner integral. Given a probability space $(\Omega,\Sigma,\mu)$ and a Banach space $(X,\parallel \cdot \parallel)$.
Def) 1)For a simple function $\Sigma_{i=1}^{n}\chi_{E_i}(\omega)x_i$ from $\Omega$ to $X$, $\int_{E}\Sigma_{i=1}^{n}\chi_{E_i}(\omega)x_id\mu=\Sigma_{i=1}^{n}\mu(E_i \cap E)x_i$
2) A function $f:\Omega\rightarrow X$ is Bochner integrable if there is a sequence of simple functions $(f_n)$ such that $lim_{n\rightarrow \infty}\int_\Omega\parallel f_n-f \parallel d\mu=0$
I wanna show $\parallel \int_E fd\mu \parallel \leq \int_E \parallel f \parallel d\mu$. This is easy for f is simple and in general, it says that an approximation guarantees the result.
I know $lim_{n\rightarrow \infty}\int_\Omega\parallel f_n-f \parallel d\mu=0$ so I can approximate (RHS). But I don't know how to approximate (LHS) of the inequality. Doesn't it use our original goal $\parallel \int_E fd\mu \parallel \leq \int_E \parallel f \parallel d\mu$? I think this is a circular reasoning