Proposition 1

Let $m$ be a vectorial measure on $T$ with values in $F$, $f$ a numerical function on $T$ that is essentially integrable for $m$. The mapping

is a linear form on $F'$.

Denoting this mapping by $\Phi$, it is immediate that $\Phi(\lambda z') = \lambda \Phi(z')$ for all $\lambda \in \mathbf{R}$. Everything comes down to showing that $\Phi(y' + z') = \Phi(y') + \Phi(z')$. Set $\mu = |y' \circ m| + |z' \circ m|$; by the Lebesgue–Nikodym theorem, one can then write $y' \circ m = g \cdot \mu$ and $z' \circ m = h \cdot \mu$, where $g$ and $h$ are two bounded and locally $\mu$-integrable numerical functions (Ch. V, §5, No. 5, Th. 2); moreover, $(y' \circ m)^+ = g^+ \cdot \mu$ and $(y' \circ m)^- = g^- \cdot \mu$, and the analogous relations hold with $y'$ replaced by $z'$ (resp. $y' + z'$) and $g$ by $h$ (resp. $g + h$). This being so, it is immediate that $f$ is essentially $\mu$-integrable (Ch. V, §2, No. 2, Cor. 1 of Prop. 3), and the relation to be proved reduces to $(g + h)^+ - (g + h)^- = (g^+ - g^-) + (h^+ - h^-)$, which is obvious.