Which numbers of [0,1) have a unique base g expansion? Good evening,
i know that is question is rather standard, but unfornunately I have not much knowledge of number theory. 
Take $2 \leq g\in \mathbb{N}$. I know that every $x \in [0,1)$ can be represented as $x = \sum_{k=1}^\infty \frac{x_k}{g^k}$ with $x_k \in \{0, .. , g-1\}$. 
I know that this repesentation is not unique (for example if $x_k= g-1 ~~ \forall k > N_0 \in \mathbb{N}$). I know that there are  $x \in [0,1)$ with a unique representation (for example if $x_k= 0 ~~ \forall k > N_1 \in \mathbb{N}$).
Here is my question: Is there any characterisation of the $x \in [0,1)$ which have a unique base g expansion?
I am very thankful for any reference! With best regards. 
 A: Here's a proof that supports the characterization stated in the comments:
The $g$-ary representations for numbers $x\in[0,1)$ are of the form 
$$\tag1 x=\sum_{k=1}^\infty x_kg^{-k}$$
with $x_k\in\{0,\ldots,g-1\}$. Let $(x_k)$ and $(y_k)$ be two distinct digit sequences in $\{0,\ldots,g-1\}$ representing the same number and wlog. $(x_k)$ lexically precedes $(y_k)$, say $x_n<y_n$ whereas $x_k=y_k$ for $k<n$.
Then
$$\begin{align}0&=\sum_{k=1}^\infty y_kg^{-k}-\sum_{k=1}^\infty x_kg^{-k}\\&=\sum_{k=1}^\infty (y_k-x_k)g^{-k}\\&=g^{-n}(y_n-x_n)+g^{-n}\sum_{k=1}^\infty (y_{n+k}-x_{n+k})g^{-k}\\&\stackrel{(1)}\ge g^{-n}+g^{-n}\sum_{k=1}^\infty (y_{n+k}-x_{n+k})g^{-k}\\&\stackrel{(2)}\ge g^{-n}+g^{-n}\sum_{k=1}^\infty(0-(g-1))g^{-k}\\
&=g^{-n}-g^{-n}(g-1)\sum_{k=1}^\infty g^{-k}
\\&=g^{-n}-g^{-n}(g-1)\frac1{g-1}=0.\end{align}$$
Since we end up with equality, we must have equality at $(1)$ and $(2)$.
But equality at $(1)$ holds if and only if $y_n=x_n+1$ and at $(2)$ if and only if for all $k\ge1$ we have $y_{n+k}=0$, $x_{n+k}=g-1$.
Note that in this case we additionally obtain
$$\tag2x = \sum_{k=1}^\infty y_kg^{-k} =\sum_{k=1}^n y_kg^{-k}=\frac{\sum_{k=1}^ny_kg^{n-k}}{g^n}.$$
Likewise, if $x=\frac m{g^n}$ in shortest trerms (i.e. with $m$ not divisible by $g$), the $g$-ary integer representation of $m$ ands in a nonzero digit, gives us a terminating representation with $y_n>0$ of $x$ via $(2)$, from which we obtain two $g$-ary representations as above.
This shows that a number $x\in[0,1)$ has two different $g$-ary representations if and only if $x=\frac m{g^n}$ with $0<m<g^n$.
