I have a question about the proof of Theorem 3.27 of Rudin's Functional Analysis. He defines vector-valued integration as follows:
Then he proves the following theorem:
You can check the statements of Theorems 3.4 and 3.20 in the end.
I could follow the steps in the proof and understand everything. However, as you can see where I marked with red color, he is arguing that the vector $m$ belongs to $\textrm{co}(K) = \textrm{co}(L(f(\mathcal{Q})))$.
My question is: could not I repeat the exact same arguments to show that $m$ actually belongs to $K$? There must be something wrong with what I am thinking since otherwise it would not be necessary to use the convex hull of $K$.
What I thought was: $K=L(f(\mathcal{Q}))$ is a compact subset of $\mathbb{R}^n$ and hence Theorem 3.4 can be applied to produce $T \in (\mathbb{R}^n)^*$ such that $Tu < Tt$ for all $u \in K$. Using the known form of the linear functionals on $\mathbb{R}^n$ (as he did), we get real numbers $c_1,\ldots,c_n$ so that $$\sum_{j=1}^nc_ju_j < \sum_{j=1}^n c_jt_j$$ for every $u=(u_1,\ldots,u_n) \in K=L(f(\mathcal{Q}))$. Hence, $$\sum_{j=1}^nc_j\Lambda_j(f(q)) < \sum_{j=1}^n c_jt_j$$ for every $q \in \mathcal{Q}$. From now on we proceed exactly as he did and conclude that $t\neq m$.
Could you help me to understand where I am going wrong?
