The finite-total-variation signed measures form a real normed space

If $\nu ,\mu \in \mathcal{M}(X,\mathcal{A})$ and $a\in \mathbb{R}$, then the [L1, L2, L4] pointwise set functions $\nu +\mu$ and $a\nu$ are again signed measures because their values are finite on every measurable set and countable additivity is preserved termwise. Also [L2] gives $\mathrm{\mid }\nu +\mu \mathrm{\mid }(X)\le \mathrm{\mid }\nu \mathrm{\mid }(X)+\mathrm{\mid }\mu \mathrm{\mid }(X),\mathrm{\mid }a\nu \mathrm{\mid }(X)=\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }\nu \mathrm{\mid }(X),$ so $\nu +\mu$ and $a\nu$ remain in $\mathcal{M}(X,\mathcal{A})$. Thus $\mathcal{M}(X,\mathcal{A})$ is closed under the pointwise operations, and the vector-space axioms are inherited from the real-valued function space on $\mathcal{A}$.

The formula $\mathrm{\parallel }\nu \mathrm{\parallel }=\mathrm{\mid }\nu \mathrm{\mid }(X)$ is nonnegative by definition. If $\mathrm{\parallel }\nu \mathrm{\parallel }=0$, [L1, L3, L4] then [L3] makes $X$ null for $\nu$, so every measurable set has $\nu$-value $0$ and therefore $\nu$ is the zero measure. Conversely the zero measure has variation $0$. Thus the separation axiom of [L4] holds.

Step 1.1 already proved absolute homogeneity and the triangle inequality: [L2, L4, step 1.1] $\mathrm{\parallel }a\nu \mathrm{\parallel }=\mathrm{\mid }a\mathrm{\mid }\mathrm{\parallel }\nu \mathrm{\parallel },\mathrm{\parallel }\nu +\mu \mathrm{\parallel }\le \mathrm{\parallel }\nu \mathrm{\parallel }+\mathrm{\parallel }\mu \mathrm{\parallel }\mathrm{.}$ Hence [L4] shows that $\mathrm{\parallel }\cdot \mathrm{\parallel }$ is a norm.

Steps 1.1 through 2.1 prove that $\mathcal{M}(X,\mathcal{A})$ is a real [step 1.1, step 1.2, step 2.1] ∎ normed space.