There is another way to attack your problem: write down a conditional that defines your set.
Consider the following set:
$$A_n := \{x\in(0,1)\mid \text{the $n$th digit of the decimal expression of $x$ is 7}\}.$$
You can see that $A_n$ is Borel, as it is a finite union of half-intervals.
Then your set can be described as follows:
$$x\in A \iff \forall n\in\mathbb{N}\exists m>n [x\in A_m].$$
Hence $A=\bigcap_{n=1}^\infty\bigcup_{m>n} A_m$, which is also Borel.
I want to remark that my strategy is a common way to determine the complexity of a given set. With some trick (by replacing $\epsilon$ or $\delta$ to $1/n$ for natural numbers $n$), you can see that the set
$$\{x\in \mathbb{R} \mid \text{$f$ is continuous at $x$}\}$$
is also Borel regardless of what $f$ is.
I found [5] in the article you mentioned particularly useful. We only need to prove a simpler version of theorem 8.1 (and corollary 8.2) in the book. So here is my answer:
Suppose $X\sim U(0,1)$. Given $b\geq 2$, and fixed $r\geq 1$. Consider the $b-$ary expansion of $x = \sum_{i=1}^\infty \sigma_ibi^{-1}$. We are interested in the probability of a given string: $(\bar{a}_1,\bar{a}_2,...\bar{a}_r)$, where $a_i = 0,1,...b-1$.
If the target string matches $x$ at index $m$, we can rewrite $x$ by replacing the digits from index $m$ to $m+r-1$:
$$
\begin{split}
x &= \sum^\infty_1 \frac{\sigma_i}{b_i}\\
&= \sum^{m-1}_{1} \frac{\sigma_i}{b^i} + \sum^{m+r-1}_{m} \frac{a_i}{b^i} + \sum_{m+r}^\infty \frac{\sigma_i}{b^i}
\end{split}
$$
Multiply both sides by $b^{m-1}$:
$$
b^{m-1}x = (b^{m-1}\sum^{m-1}_{1} \frac{\sigma_i}{b^i}) + \frac{a_mb^{r-1}+a_{m+1}b^{r-2}+ ...a_{m+r-1}}{b^r} + \sum_{r+1}^\infty \frac{\sigma_{m+i-1}}{b^i}
$$
The fractional part $\{b^{m-1}x\}$ doesn't contain $(b^{m-1}\sum^{m-1}_{1} \frac{\sigma_i}{b^i})$, and we have $\sum_{r+1}^\infty \frac{\sigma_{m+i-1}}{b^i}< 1b^{-r}$ since they are the trailing digits of this expansion. Therefore,
$$
\{b^{m-1}x\}\in [\frac{a_mb^{r-1}+a_{m+1}b^{r-2}+ ...a_{m+r-1}}{b^r}, \frac{a_mb^{r-1}+a_{m+1}b^{r-2}+ ...a_{m+r-1} + 1}{b^r}) = I_r
$$
Whenever there is a matching sequence, we have $\{b^{m-1}x\}\in I_r$.
Since $X\sim U(0,1)$, we have that $\{b^{m-1}X\}\sim U(0,1)$ because any fractional part is uniformly distributed across each interval $[n,n+1)$, where $n = 0,1,...b^{m-1}-1$. Since $b^{m-1}X$ is uniform, each interval has probability $1/b^{m-1}$ and therefore the marginal distribution of $\{b^{m-1}X\}$ is uniform $(0,1)$.
Given the first $N$ digits of the expansion of $x$, we need to check $N-r+1$ possible starting point of the desired sequence $b_1b_2...b_r$ and this is equivalent to $\sum^{N-r+1}_1X1_{I_r}$, where $X\sim U(0,1)$. Hence by SLLN,
$$
\lim_n \frac{\{\# k = 1,2,...n: (\sigma_k...\sigma_{k+r}) = (b_1,b_2...b_r)\}}{n} = \lim \frac{\sum^{n-r+1}_1X1_{I_r}}{n-r+1}\frac{n-r+1}{n}\xrightarrow[]{}\frac{1}{b^r}
$$
almost surely.
This is true for any $r\geq 1$, so $X$ is normal almost surely. We can do the same for every $b\geq 2$, and then take the countable intersection of the above events so that $X$ is normal with respect to every base $b\geq 2$ with probability 1. We are done.
See theorem 8.1 in https://web.maths.unsw.edu.au/~josefdick/preprints/KuipersNied_book.pdf
Best Answer
Yes, this is a Borel set.
It is often quite difficult to prove something is a Borel set by means of “mapping in” to the set. Your set is defined as the image of a certain measurable set. The trick is to instead define it as the inverse image of a certain measurable set.
More formally, define, for all $x \in [0, 1]$ and $n \in \mathbb{N}_+$, $x_n \in \{0, 1\}$ to be the $n$th bit of $x$’s binary expansion after the radix point. That is, we have $x = \sum\limits_{n = 1}^\infty x_n 2^{-n}$. It turns out not to matter how you define the expansion of a dyadic number (one with a terminating and a nonterminating expansion),so just pick one definition and stick with it.
It is fairly easy to show that each function $x \mapsto x_i$ is Borel measurable. Now define $f_i(x) = \frac{x_1 + \cdots + x_i}{i}$, which is also clearly measurable. Then $S = \{x \in [0, 1] \mid \lim\limits_{n \to \infty} f_n(x) = 1/2\}$ (check this for dyadic numbers). Note that we redefine $S$ by mapping out of $[0, 1]$, not by mapping in.
By expanding the definition of $\lim\limits_{n \to \infty}$, we see that $S = \bigcap\limits_{q \in \mathbb{Q}_+} \bigcup\limits_{n \in \mathbb{N}_+} \bigcap\limits_{m \in \mathbb{N}, m \geq n} f_m^{-1}(B_q(1/2))$, where $B_q$ is a ball of radius $q$ as usual. Using measurability and the properties of a $\sigma$-algebra, $S$ is Borel.