Quantum state tomography owes its power and flexibility to the fact that it supports a wide class of measurements. Any informationally complete POVM, i.e. one whose elements span the space $L_H(\mathcal{H})$ of Hermitian operators on the target system's Hilbert space $\mathcal{H}$ qualifies for use in QST.
One way to highlight the generality of QST with respect to the choice of POVM is to cast it as a vector reconstruction problem. Every orthonormal basis $\{v_k\}$ of a vector space $V$ has the reconstruction property that $u = \sum_k \langle u, v_k\rangle v_k$ for any vector $u$. It turns out that certain sets of vectors other than bases, namely frames, also have a variant of the property. More precisely, for any vector $u$ and a frame $\{v_k\}$ we have
$$u = \sum_k \langle u, v_k\rangle\tilde{v_k}\tag1$$
where $\{\tilde{v_k}\}$ is a frame dual to $\{v_k\}$. The dual frame can be computed as $\tilde{v_k} = S^{-1}v_k$ where $S$ is the frame operator defined as $S: u \mapsto \sum_k\langle u, v_k\rangle v_k$.
Quantum state tomography can be analyzed as an application of frame theory to the task of reconstructing an element of $L_H(\mathcal{H})$. Suppose that positive operators $E_k$ sum to identity and span $L_H(\mathcal{H})$, i.e. $\{E_k\}$ is both a POVM and a frame. Let $\{F_k\}$ denote the frame dual to $\{E_k\}$. Then by $(1)$ for any operator $\rho$ we have
$$
\rho = \sum_k \mathrm{tr}(\rho E_k)F_k.\tag2
$$
This enables complete characterization of a quantum state, because the coefficients $\mathrm{tr}(\rho E_k)$ are accessible experimentally while operators $F_k$ can be derived using frame theory summarized above.
The key point about this construction is that any POVM that is also a frame qualifies for use in quantum state tomography, so we need to make additional choices to exhibit a specific POVM.
Example 1: Standard QST
Any orthonormal basis is a frame which is its own dual. A well-known example of an orthonormal basis in $L_H(\mathbb{C}^2)$ is the set $\{I/\sqrt{2}, X/\sqrt{2}, Y/\sqrt{2}, Z/\sqrt{2}\}$. In this case equation $(2)$ takes the form
$$
\rho = \frac12\left(\mathrm{tr}(\rho)I + \mathrm{tr}(\rho X)X+ \mathrm{tr}(\rho Y)Y + \mathrm{tr}(\rho Z)Z\right).
$$
However, the Pauli operators are not positive and hence not a POVM. We can obtain a POVM by replacing each operator in the set with its two eigenprojectors and renormalizing
$$
\left\{\frac{|0\rangle\langle 0|}{3}, \frac{|1\rangle\langle 1|}{3},
\frac{|+\rangle\langle +|}{3}, \frac{|-\rangle\langle -|}{3},
\frac{|{+i}\rangle\langle {+i}|}{3}, \frac{|{-i}\rangle\langle {-i}|}{3}, \right\}.
$$
This set spans $L_H(\mathbb{C}^2)$, because the original orthonormal basis does and is therefore an informationally complete POVM. The advantage of this POVM is that it is often relatively simple to realize experimentally.
Example 2: Minimal POVM
A frame spans its vector space, so the smallest informationally complete POVM has at least $\dim L_H(\mathbb{C}^2) = 4$ elements. There are many examples that attain this minimum. For instance, define
$$
\begin{align}
|\psi_0\rangle &= |0\rangle \\
|\psi_1\rangle &= \frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac23}|1\rangle \\
|\psi_2\rangle &= \frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac23}e^{\frac{2\pi i}{3}}|1\rangle \\
|\psi_3\rangle &= \frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac23}e^{\frac{4\pi i}{3}}|1\rangle
\end{align}
$$
and set $E_k = \frac12|\psi_k\rangle\langle\psi_k|$ (see also this picture). Following standard frame theory we calculate
$$
\begin{align}
F_0 &= \begin{pmatrix}
2 & 0 \\
0 & -1
\end{pmatrix} \\
F_1 &= \begin{pmatrix}
0 & \sqrt{2} \\
\sqrt{2} & 1
\end{pmatrix} \\
F_2 &= \begin{pmatrix}
0 & -\frac{1+i\sqrt{3}}{\sqrt{2}} \\
-\frac{1-i\sqrt{3}}{\sqrt{2}} & 1
\end{pmatrix} \\
F_3 &= F_2^T
\end{align}
$$
which together with $E_k$ and equation $(2)$ provides a complete description for single-qubit QST using a minimal POVM.
