 Solution

For one fixed POVM repeated on independently prepared copies, the Born rule gives $m$ outcome probabilities with one normalization constraint. An arbitrary mixed qubit density operator has three independent real parameters,The probability map is affine in $\mathbf{r}$. It can be injective on the full Bloch ball only if $m-1\ge 3$. If its rank is smaller, a nonzero undetected direction gives two nearby physical Bloch vectors with the same probabilities. Hence $m\ge 4$ for an informationally complete POVM.

To attain this bound, take the four directionsThey are the vertices of a regular tetrahedron. The tetrahedral qubit POVM has effects $E_i=(I+{\mathbf{n}}_i\cdot \mathbf{\sigma })/4$, each positive and rank one, and ${\sum }_i{E}_i=I$. The observed probabilities areSince ${\sum }_i{\mathbf{n}}_i=0$ and ${\sum }_i{\mathbf{n}}_i{\mathbf{n}}_i^T=4I_3/3$, they recover the state byThis is the minimum for a single fixed measurement family. If the wording permits changing measurement settings between groups of copies, three binary projective measurements of the three Pauli observables also suffice, each with $m=2$. That procedure is informationally complete three-observable qubit tomography, and it does not contradict the four-outcome bound for one fixed POVM. Tomography estimates probabilities statistically from many copies; a finite sample does not determine an arbitrary continuous state exactly.