Let $g:\mathbb{R}\to\mathbb{R}$ and $h:\mathbb{R}\to\mathbb{R}$ be such that $h(x) = \operatorname{sgn}(g(x))$.

Recall: $\operatorname{sgn}(t) = \begin{cases} 1, & t>0\\ 0, & t=0\\ -1, & t<0 \end{cases}$

Check each statement:

1) "The domain of $h(x)$ is the same as the domain of $g(x)$." $\Rightarrow$ True, because $\operatorname{sgn}(g(x))$ is defined for every $x$ where $g(x)$ is defined.

2) "The domain of continuity of $h(x)$ equals the domain of continuity of $g(x) - \{x\in\mathbb{R} : g(x)=0\}$."

At points where $g(x)\neq 0$, $h(x)$ is locally constant ($1$ or $-1$), hence continuous there (provided $g$ itself is continuous).

At points where $g(x)=0$, $h(x)$ jumps from $-1$ to $1$, so it is discontinuous. $\Rightarrow$

This statement is true.

3) "The domain of $h(x)$ is different from the domain of $g(x)$ at the same point."

Since for every $x$ in the domain of $g$, $h(x)=\operatorname{sgn}(g(x))$ is defined, the domains are exactly the same; they never differ.

$\Rightarrow$ This statement is false.

4) " $h(x)$ is discontinuous at $g(x)=0$."

At any $x_0$ where $g(x_0)=0$, the left and right limits of $h(x)$ are $-1$ and $1$, not equal to $h(x_0)=0$.

$\Rightarrow$ $h$ is discontinuous there, so this statement is true.

Therefore, the statement which is **not true** is: $\boxed{\text{Option 3}}$

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