Null functions form a linear subspace and are exactly the zero-seminorm class

Fix $1\le p<\mathrm{\infty }$. If $f,g\in {\mathcal{N}}_p(\mu )$, choose measurable null sets $E_f,E_g$ outside which $f=0$ and $g=0$. Their union is null, and on its complement one has $f+g=0$ and $af=0$ for every $a\in \mathbb{R}$. Because ${\mathcal{L}}^p(\mu )$ is a vector space, [L4] makes ${\mathcal{N}}_p(\mu )$ a linear subspace.

If $f\in {\mathcal{N}}_p(\mu )$, then $\mathrm{\mid }f{\mathrm{\mid }}^p=0$ almost everywhere, so [L1, L5] $\int \mathrm{\mid }f{\mathrm{\mid }}^{p}d\mu =0,$ hence $\mathrm{\parallel }f{\mathrm{\parallel }}_p=0$. Conversely, if $\mathrm{\parallel }f{\mathrm{\parallel }}_p=0$, then the same theorem [L5] forces $\mathrm{\mid }f{\mathrm{\mid }}^p=0$ almost everywhere and therefore $f=0$ almost everywhere.

If $f\in {\mathcal{N}}_{\mathrm{\infty }}(\mu )$, then $0$ is an essential bound for $\mathrm{\mid }f\mathrm{\mid }$, so $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}=0$. Conversely, if $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}=0$, then [L6] gives $\mathrm{\mid }f\mathrm{\mid }\le 0$ almost everywhere, hence $f=0$ almost everywhere.

If $f,g\in {\mathcal{N}}_{\mathrm{\infty }}(\mu )$, the same null-set union argument as in step 1.1 shows that $f+g$ and $af$ vanish almost everywhere, and [L2] places them in $L^{\mathrm{\infty }}(\mu )$. Therefore [L4] makes ${\mathcal{N}}_{\mathrm{\infty }}(\mu )$ a linear subspace.

Steps 1.1 and 2.1 prove the two subspace claims, and steps 1.2 and 1.3 identify the zero-seminorm class in every range.