# Disprove: for an upper semi-continuous function $g$, the set $\{x \in \mathbb{R}^n: g(x) \leq 0\}$ is closed

I need to find a counterexample to disprove the following statement:

for upper semi-continuous function $g: \mathbb{R}^n \rightarrow \mathbb{R}$, the set $\{x \in \mathbb{R}^n: g(x) \leq 0\}$ is closed.

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For the case of n=1 you can take the function $f(x)=x$ when $x<0$ and $f(x)=1+x$ when $x\geq 0$ then it is continuous everywhere except in $x=0$ where it is upper semi continuous. The set $f^{-1}(y\leq 0)= (-\infty,0)$ which is not close.
you can extend this for every n, by taking a function which is continuous everywhere except a hyper plane (which in n=1 is just a point). So in n=2 do the same thing just the jump will be on the line $x=0$