Suppose Alice gives Bob a qubit prepared in one of two states:

$|\psi_1\rangle = |0\rangle \qquad \text{or} \qquad |\psi_2\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle).$

Here is a measurement strategy for Bob to determine unequivocally whether he receives $|\psi_1\rangle$ or $|\psi_2\rangle$. Define:

$\mathbf{E}_1\triangleq\dfrac{\sqrt{2}}{1+\sqrt{2}}|1\rangle\langle1|, \quad \mathbf{E}_2\triangleq\dfrac{\sqrt{2}}{2(1+\sqrt{2})}(|0\rangle-|1\rangle)(\langle0|-\langle1|), \qquad \mathbf{E}_3\triangleq\mathbf{I}-\mathbf{E}_1-\mathbf{E}_2.$

Clearly $\mathbf{E}_i$ satisfy the completeness relation, and by checking the definition of positive semidefiniteness, we can verify $\mathbf{E}_i$ to be positive operators, so $\mathbf{E}_i$ form a POVM.

Let us now see how Bob can use $\mathbf{E}_i$ to distinquish between $|\psi_1\rangle$ and $|\psi_2\rangle$:

• Since $\Pr\{1|\psi_1\}=\langle\psi_1|\mathbf{E}_1|\psi_1\rangle=0$ whereas $\Pr\{1|\psi_2\}=\langle\psi_2|\mathbf{E}_1|\psi_2\rangle\neq0$, getting a measurement outcome associated with $\mathbf{E}_1$ implies Bob must have received $|\psi_2\rangle$.
• Since $\Pr\{2|\psi_1\}=\langle\psi_1|\mathbf{E}_2|\psi_1\rangle\neq0$ whereas $\Pr\{2|\psi_2\}=\langle\psi_2|\mathbf{E}_2|\psi_2\rangle=0$, getting a measurement outcome associated with $\mathbf{E}_2$ implies Bob must have received $|\psi_1\rangle$.
• Receiving a measurement outcome associated with $\mathbf{E}_3$ however precludes Bob from inferring anything about the identity of the state he receives.

Using $\mathbf{E}_i$, Bob never mistakes $|\psi_1\rangle$ for $|\psi_2\rangle$ and vice versa, but Bob sometimes cannot determine which state he receives.

Denote by $\{|i\rangle\}_i$ an orthonormal basis.

Define $\mathbf{E}_i\triangleq|i\rangle\langle i|$ (a projector), and $\mathbf{\sigma}\triangleq\sum_i\mathbf{E}_i$, then for each $j$,

$\sigma|j\rangle = |j\rangle \implies (\sigma - \mathbf{I})\mathbf{A} = \mathbf{0},$

where $\mathbf{A}$ is a matrix with the basis vectors $\{|i\rangle\}_i$ as columns. The linear independence of the basis vectors implies $\mathbf{A}$ is nonsingular, and

$\sigma - \mathbf{I} = \mathbf{0}\mathbf{A}^{-1} \implies \sigma = \mathbf{I} \implies \sum_i\mathbf{E}_i = \mathbf{I}.$

In other words, $\mathbf{E}_i$ satisfies the completeness relation.

Given a quantum state $|\psi\rangle = \sum_i\alpha_i|i\rangle$, we should expect $\Pr\{i|\psi\} = |\alpha_i|^2$, and indeed

$\Pr\{i|\psi\} = \langle\psi|\mathbf{E}_i|\psi\rangle = \alpha_i^\ast\langle i| \cdot |i\rangle\langle i| \cdot \alpha_i|i\rangle = |\alpha_i|^2 \geq 0.$

At this point, we can call $\{\mathbf{E}_i\}$ a POVM.