Minimum with Lagrangian why we only consider $\lambda$ in the set $ Y = \{ \lambda : \min \space L ( x, \lambda ) < - \infty \} $

Question

I'm reading these lecture notes about optimization:

http://www.statslab.cam.ac.uk/~rrw1/opt/O.pdf http://dcs.gla.ac.uk/~fischerf/teaching/opt/notes/notes.pdf

and they show the same steps to find the minimum:

1. For each $\lambda$ solve the problem

minimize $L(x, \lambda)$ subject to $x \in X$.

2. Define the set $ Y = \{ \lambda : \min \space L ( x, \lambda ) < - \infty \} $

3. For $\lambda \in Y$ , the minimum will be obtained at some $x(\lambda)$ (that depends on $\lambda$ in general).

4. Adjust $\lambda \in Y$ so that $x(\lambda)$ is feasible. If $\lambda^* \in Y$ exists such that $x^* = x(λ^∗)$ is feasible then $x^∗$ is optimal for the problem.

Is that step necessary? Why should we exclude them? What if the minimum is at $-\infty$?