How can you find the continuous digits of $g(x)=3 \lfloor x \rfloor^3$?

I have a task that I'm just not getting anywhere with:

Let $ g: \mathbb{R} \rightarrow \mathbb{R} $ given by $g(x)=3 \lfloor x \rfloor^3$ for all $ x \in \mathbb{R} $.
Determine all places where $g$ is continuous. Justify your assertions in the process.

From the picture you can see that the function is not continuous on the whole of R, but only on these single horizontal areas. If my intuition is correct, this function should be continuous at all points $a \in \mathbb{R} \setminus \mathbb{Z}$ and discontinuous at points $a \in \mathbb{Z}$.

But how can I now show these two cases?

I found a proof of the normal floor function in an old script:

Is it possible to do the proof like this for my given function?

(f) The function $ \mathbb{R} \rightarrow \mathbb{R}, x \mapsto\lfloor x\rfloor $ is continuous at all points $ a \in \mathbb{R} \backslash \mathbb{Z} $ and discontinuous (i.e., not continuous) at the points $ a \in \mathbb{Z} $.

Proof. For let $ a \in \mathbb{Z} $. Then for $ x_{n}=a+\frac{1}{2 n} $ that $ x_{n} \rightarrow a $ and $ \left\lfloor x_{n}\right\rfloor=a $, also $ \lim \limits_{n \rightarrow \infty}\left\lfloor x_{n}\right\rfloor=a $. However, for $ x_{n}=a-\frac{1}{2 n} $ that $ x_{n} \rightarrow a $ and $ \left\lfloor x_{n}\right\rfloor=a-1 $, also $ \lim \limits_{n \rightarrow \infty}\left\lfloor x_{n}\right\rfloor=a-1 $. The limit also does not exist for $ a \in \mathbb{Z} $.

Let $ a \in \mathbb{R} \backslash \mathbb{Z} $. Then $ \lfloor a\rfloor<a<\lfloor a+1\rfloor $ holds. Moreover, $ \lfloor x\rfloor=\lfloor a\rfloor $ holds for all $ x \in(\lfloor a\rfloor,\lfloor a+1\rfloor) $, which is an open interval containing $ a $. If $ x_{n} \rightarrow a $ also holds for $ n \rightarrow \infty $, then $ x_{n} $ is finally contained in $ (\lfloor a\rfloor, \lfloor a+1\rfloor) $. Thus $ \lim \limits_{n \rightarrow \infty}\left\lfloor x_{n}\right\rfloor=\lfloor a\rfloor $ follows. This shows continuity in $ a $.

Solution 1:

If $x$ is an integer, then $\lfloor x-\delta\rfloor=\lfloor x\rfloor-1$ for all $0<\delta<1$. So you can't make $|f(x-\delta)-f(x)|<\epsilon$ and the function is discontinuous.

On the other hand, if $x$ is not an integer, when $\delta<\min(x-\lfloor x\rfloor, \lfloor x\rfloor+1-x)$, $f(x)$ is constant, hence continuous.

Illustration: