Composition of a vector measure and a linear functional

Question

Assume that $X$ is topological vector space over the field $\mathbb{R}$. Let $\mathcal{M}$ be a $\sigma$-algebra of subsets of a nonempty set $T$. We say that $$m: \mathcal{M}\to X$$ is a vector measure if $$m(E_1\cup E_2)=m(E_1)+m(E_2)$$ whenever $E_1$ and $E_2$ are disjoint members from $\mathcal{M}$.

Question. Why is it that for each $x'\in X'$, the composition map $$x'\circ m: \mathcal{M}\to \mathbb{R}$$ is a measure on $\mathcal{M}$ whenever the vector measure $m:\mathcal{M}\to X$ is countably additive?

In the above, $X'$ denotes the space of all continuous linear functionals on $X$. So far, I can show that $(x'\circ m)(\varnothing)=0$. What bothers me is to show that $x'\circ m$ is countably additive. Any tips? Thanks in advance...

Answer 1

Let $\{E_i:i\in \mathbb{N}\}\subseteq \mathcal{M}$ where the $E_i's$ are pairwise disjoint. Since $m:\mathcal{M}\to X$ is countably additive, $$m\left(\bigcup_{i=1}^{+\infty}E_i\right)=\sum_{i=1}^{+\infty}m(E_i).$$ Because $x'\in X'$, as $n\to +\infty$ we get $$\sum_{i=1}^{n}(x'\circ m)(E_i)=x'\left(\sum_{i=1}^{n}m(E_i)\right)\to x'\left(\sum_{i=1}^{+\infty}m(E_i)\right)=(x'\circ m)\left(\bigcup_{i=1}^{+\infty}E_i\right).$$ Hence $$\sum_{i=1}^{+\infty}(x'\circ m)(E_i)=(x'\circ m)\left(\bigcup_{i=1}^{+\infty}E_i\right).$$ This proves that $x'\circ m$ is countably additive.

Answer 2

You should double check the meaning of linear functionals to see the answer to your question. $m$ is countable additive, so $x$ applied to a linear combination gives....