I've explained quantum measurements more formally in my other answer, but let me just reproduce what you need here.
There is a difference between applying some operator and measuring a state. A measurement of some operator $A$ on some system in the state $\left| \psi \right\rangle$ can be expressed if we know the eigenvalues of $A$ (which are also the possible outcomes upon measurement) and corresponding eigenvectors of $A$.
Supposing $A$ has $M$ unique eigenvalues labelled $a^{\,}_m$, I can write $A$ as
$$ A \, = \, \sum\limits_{m=0}^{M-1} \, a^{\,}_m \, \mathbb{P}^{\,}_m \, ,$$
where the projectors $\mathbb{P}$ sum to the identity, square to themselves, and are orthogonal to each other.
If there is only one eigenvector $\left| \phi_m \right\rangle$ for the eigenvalue $a^{\,}_m$, then $\mathbb{P}^{\,}_m = \left| \phi_m \middle\rangle \hspace{-0.3mm} \middle\langle \phi_m \right|$. If there are multiple eigenvectors $\left| \phi^{k}_m \right\rangle$ for eigenvalue $a^{\,}_m$, then $\mathbb{P}^{\,}_m = \sum_k \left| \phi^k_m \middle\rangle \hspace{-0.3mm} \middle\langle \phi^k_m \right|$.
The possible measurement outcomes are the unique $a^{\,}_m$. The corresponding probability (for outcome $a^{\,}_m$) is just $p^{\,}_m = \left\langle \psi \middle| \mathbb{P}^{\,}_m \middle| \psi \right\rangle$. Technical explanation below:
Upon measuring $A$ on a system in the state $\left| \psi \right\rangle$, the new state is
$$ \left| \psi \right\rangle \, \to \, \left| \psi' \right\rangle \, = \, \sum\limits_{m=0}^{M-1} \, \left( \mathbb{P}^{\,}_m \, \left| \psi \right\rangle \right)^{\,}_{\rm ph} \otimes \left| m \right\rangle^{\,}_{\rm meas} \, ,$$
where the state labelled "ph" is the post-measurement state of the physical system, and the state labelled "out" records the outcome—it is the state of the measurement apparatus if you like.
Note that $\mathbb{P}^{\,}_m \, \left| \psi \right\rangle$ is the unnormalized Copenhagen wavefunction following measurement, given that outcome $m$ obtained.
The probability of getting outcome $m$ is
$$p^{\,}_m \, = \, \left\langle \psi' \middle| \mathbb{1} \otimes \widetilde{\mathbb{P}}^{\,}_{m} \middle| \psi' \right\rangle \, ,$$
where the tilde denotes an operator on the apparatus. The probability of outcome $m$ is just the expectation value that the detector is in the state $m$ after measurement.
Anyway, if you want to measure $L^{\,}_z$, you just need to work out the projectors onto its eigenstates in some basis. If you're already working in the $L^{\,}_z$ eigenbasis, then it's easier. But the possible measurement outcomes are always the eigenvalues, and the probability of an outcome is the square of the overlap of the pre-measurement state with the eigenstate corresponding to that outcome. If there are multiple such states, sum over each individual overlap squared.
$$ p^{\,}_m = \left| \left\langle \phi^{\,}_m \middle| \psi \right\rangle \right|^2~~~\text{or}~~~p^{\,}_m = \sum_k \left| \left\langle \phi^{k}_m \middle| \psi \right\rangle \right|^2\,,$$
for nondegenerate $m$ and degenerate $m$, respectively.
