Bochner integral norm inequality

Restrict a defining simple approximation to $E$. [given, L1, choose] Choose integrable simple $s_n$ with $\int \mathrm{\parallel }f-s_n\mathrm{\parallel }\to 0$. Replacing each by ${\mathbf{1}}_E{s}_n$ shows from [L1] that ${\mathbf{1}}_{E}f$ is Bochner integrable and that its integral is the norm limit of ${\int }_E{s}_n$.

Bound each simple integral. [L2, L3, step 1.1] The triangle inequality and [L2] give $\mathrm{\parallel }{\int }_E{s}_n\mathrm{\parallel }\le {\int }_E\mathrm{\parallel }{s}_n\mathrm{\parallel }$. Since $\mathrm{\parallel }{s}_n\mathrm{\parallel }\le \mathrm{\parallel }f\mathrm{\parallel }+\mathrm{\parallel }{s}_n-f\mathrm{\parallel }$, [L3] yields $\mathrm{\parallel }{\int }_E{s}_n\mathrm{\parallel }\le {\int }_E\mathrm{\parallel }f\mathrm{\parallel }+{\int }_E\mathrm{\parallel }{s}_n-f\mathrm{\parallel }$.

Pass to the limit and conclude. [step 1.1, step 2.1] Let $n\to \mathrm{\infty }$ in step 2.1. Norm continuity and step 1.1 identify the left limit with $\mathrm{\parallel }{\int }_{E}f\mathrm{\parallel }$, while the last term tends to zero. This proves the inequality, including $E=\mathrm{\varnothing }$ and $f=0$.