Will perturbation function be continuous?

Question

Let's consider minimization problem in the standard form i.e: $$\min_{g(x)\leq 0,\:x\in\mathbb{X}}f(x)$$

Let's definie perturbation function on the set $D_M=\{z: W_z= \{x: g(x)\leq z\}\:\text{is non-empty}\:\}$

$$M(z)=\inf_{g(x)\leq z,x\in\mathbb{X}}f(x)$$

So my question is:

Let's assume that $f$ and $g$ are both continuous, does it follow that $M$ is continuous? I tried to come up with counterexamples, which proved to be difficult, so I think this might be true. Any help appreciated.

Answer 1

A counterexample:

Let $\mathbb{X} = \mathbb{R}$, $f(x) = -x$, $g(x) = (x+1)x(x-2)^2$. Then $M(0) = -2$, but $M(-\varepsilon) \geq 0$ for any $\varepsilon > 0$.

You can find sufficient conditions for continuity of the value function here: https://www2.isye.gatech.edu/people/faculty/Alex_Shapiro/SIAM-R.pdf