@Jakobian My idea is basically this: $(\tilde \psi \circ \tilde \sigma^{-1})(\varphi) = \tilde \psi(\tilde \sigma^{-1}(\varphi)) = \tilde \psi(\varphi \otimes 1_V) = \varphi(1_V)$ but I'm not sure what this $1_V$ exactly is, haha
@Jakobian Does this approach look right to you, i.e. if we pick some 'good' $1_V$, could we get the trace with this?
Consider the map $\varphi:\sum a_iv_i\mapsto a_jv_k$. Those maps form a basis of $\text{Hom}(V, V)$. Then $\tilde{\sigma}^{-1}(\varphi) = \pi_j \otimes v_k$ where $\pi_j(\sum a_i v_i) = a_j$. So $\tilde{\psi}(\pi_j \otimes v_k) = \pi_j(v_k) = \delta_{jk} = \text{trace}(\varphi)$.
Since both of these maps are linear, they must be equal
@psie the outer measure is such that the $\mu^*$-measurable sets include the algebra $\mathcal{A}$. So they also include their countable intersections, and countable unions of their countable intersections.
@Jakobian I'm reviewing this stuff, so I feel a bit forgetful :) this is Caratheodory's theorem, that the $\mu^\ast$-measurable sets form a $\sigma$-algebra and the restriction of the outer measure to this $\sigma$-algebra is a complete measure.
@Jakobian Thank you, that's elegant. Is there also a more direct way - without considering these special $\varphi$ first - right away doing it in full generality?
We can also consider the general $\varphi: \sum a_i v_i \mapsto \sum b_i v_i$ right away, then we just need to 'pull apart' the tensor product and basically argue analogously, right?
