Let $\mathcal{M}$ be a $\sigma$-algebra of subsets of a nonempty set $T$, $X$ is Hausdorff LCTVS, $X'$ the topological dual of $X$, and $m: \mathcal{M}\to X$ a countably additive vector measure on $\mathcal{M}$.
Let $f: T\to \mathbb{R}$ be $\mathcal{M}$-measurable function. We say that $f$ is $m$-integrable if for each $x'\in X'$, $f$ is integrable with respect to the measure $x'\circ m:\mathcal{M}\to [0,+\infty]$ and for each $E\in \mathcal{M}$ there exists $x_E\in X$ such that for any $x'\in X'$, we have $$x'(x_E)=\int_Ef\ d(x'\circ m).$$ We shall denote $x_E=\int_Ef\ dm.$
Suppose that $f$ is $m$-integrable. Define $$n:\mathcal{M}\to X$$ by $$n(E)=\int_Ef\ dm$$ for each $E\in \mathcal{M}.$ I need to show that the mapping $n$ is well-defined. How do we do this?
Suppose $x_E$ has the property $(\ast)$ and $y \neq x_E$. Since $X$ is Hausdorff, the Hahn-Banach theorem ensures that there is $x' \in X'$ such that $x'(y) \neq x'(x_E)$. But then $$ x'(y) \neq x'(x_E) = \int_{E} f\,d(x'\circ m) $$ shows that $y$ can't have the property $(\ast)$.
The definition of $m$-integrability of $f$ thus implies the existence of a unique $x_E$ having the property in $(\ast)$ and the map $$ \begin{align*} n \colon \mathcal{M} & \longrightarrow X \cr E & \longmapsto x_E \end{align*} $$ is well-defined. The point $x_E$ is the definition of the integral $x_E = \int_{E} f\,dm$.