(Apologies for the poor typesetting of kets. I haven't been able to figure out how to do that on this site).
On page 85 of Nielsen and Chuang's textbook, they write that the probability of obtaining the result $m$ after experiment $M_m$ conducted on qubit $\psi$ is given by
\begin{equation} p(m) = \langle \psi \mid M_m^\dagger M_m\mid \psi\rangle \end{equation}
This is the first they said about measurements, so I'll take their word for this; no problem. My confusion comes later down the page, when they write that the state after the measurements $M_0$ and $M_1$, respectively, are
\begin{align} \frac{M_0\mid\psi\rangle}{\mid a\mid} &= \frac{a}{\mid a\mid}\mid0\rangle \\ \frac{M_1\mid\psi\rangle}{\mid b\mid} &= \frac{b}{\mid b\mid}\mid1\rangle \end{align}
In the first case, with $M_0$: as long as $\mid a \mid \neq 0$, then the qubit will be guaranteed to be in state $\mid 0 \rangle$. Similarly, if $\mid b \mid \neq 0$ then the qubit is guaranteed to be in state $\mid 1\rangle$.
This doesn't seem like a measurement to me. All that's being measured is whether or not the amplitude for one of the basis vectors is nonzero.
For example, if you have the state $\psi \equiv \left(\mid0\rangle + \mid1\rangle\right)/\sqrt{2}$ , then using the measurement $M_0$ will always result in the qubit being in state $\mid 0\rangle$, and measurement $M_1$ always leads to $\mid 1 \rangle$. This doesn't seem very useful.
What am I missing here?
Editing to add: Moreover, these measurement operators aren't even applicable in the special case where the qubit is just a classical bit (one of the amplitudes equal to one). For instance, how can you even apply the operator $M_0 = \mid 0 \rangle \langle 0 \mid$ to the state $\mid 1 \rangle$? This would just give the state $0\mid 0 \rangle + 0\mid 1 \rangle$, which makes no sense.
