I just built a non-measurable set without the axiom of choice, where is my mistake? Suppose $p \in \mathbb{R}$. Let us define $p_k$ the $k\text{-th}$ decimal. Let us define, whet it exists, the limit of the mean of the $k\text{-th}$ first decimal.
$$c_p = \lim_{n \rightarrow \infty} \frac{\sum_{1}^{n}{p_k}}{n}$$
Let us define
$$R_5 = \{r \in [0, 1] | c_r = 5\}$$
$R_5$ is not empty, since $0.55555... \in R_q$.
It is also dense in $[0,1]$. (if $x < y$, you can find the first different decimal, and replace the end with 55555... on either $x$ or $y$, depending on their next decimal : e.g. if $x=0.4566666..$, $y=0.457777...$, choose $0.457555...$).
$R_5$ is uncountable, since you can use the Cantor diagonalization process, and build the $q_k$ decimal as 5 if the $q_k$ decimal of the number it is compare to is not a 5, and alternate between 1 and 9 if it is a 5.
$R_5^{c}$ is also uncountable and dense in $[0, 1]$, and even in every $[a, b]$, with $0 \le a < b \le 1$.
Since I can build some $R_p$ for each $p \in ]1, 8[$ using a similar trick, and they are trivially disjoint, $\mu(R_p) = 0$ is the only possible solution. But since it is dense and uncountable, I cannot apply a similar trick as what is used to measure the Cantor's set or the rational set.
It seems like $R_5$ is not Lebesgue-measurable (the outer Lebesgue measure is not equal to the inner Lebesgue measure), and was designed without the use of the axiom of choice. But, as said in a lot of answer, this is impossible (with the ZF axioms alone, every set of $\mathbb{R}$ is measurable), see e.g. https://en.wikipedia.org/wiki/Non-measurable_set. So, where is my mistake ?
 A: $R_5$ is a Borel set, and therefore Lebesgue measurable. Define $g_n(x)=\frac1n\sum_{k=1}^n\left\lfloor10^kx\right\rfloor-10\lfloor 10^{k-1}x\rfloor$ and consider $u(x)=\limsup_{n\to\infty}g_n(x)$ and $v(x)=\liminf_{n\to\infty} g_n(x)$. These are Borel functions because they are the liminf and limsup of Borel functions, and $R_5=v^{-1}\{5\}\cap u^{-1}\{5\}$.
A: User Mees de Vries (if I recall correctly) had posted an answer, with some minor errors, that explained Lebesgue measurability of $R_5$ in terms of normal numbers. Since it's been a day since and since that approach allows the actual computation of the measure of $R_5$ and many similar sets, I'm posting as community wiki a version with the corrections needed and a few sources.
It is known (here's a proof by by Edmund Weitz, based on an idea by Hausdorff; to my understanding, it works just fine in ZF) that almost all real numbers, in the sense of Lebesgue measure, are normal.
Now let $N$, $N_b$ and $S_b$ denote the sets of normal, base-$b$ normal and base-$b$ simply normal numbers, and call $R^b_p$ the same construction you've done, but in base $b$. The following inclusions are easily proved: $$R^b_{(b-1)/2}\supseteq[0,1]\cap S_b\supseteq[0,1]\cap N_b\supseteq[0,1]\cap N$$
Therefore we know, for instance, that $R^{10}_{4.5}$ has Lebesgue measure $1$, and therefore that $R^{10}_5$ is a null set. More generally, all the $R^b_p$ are Lebesgue measurable and their measure is $1$ if $p=\frac{b-1}2$ and $0$ otherwise.
If we consider that $\bigcap_{b\ge2}R^b_{(b-1)/2}\supseteq[0,1]\cap N$, we can see that any asymptotic condition, no matter how strange it may look, on the average of the digits of a number ends up producing a null set, and therefore a Lebesgue measurable one, as soon as in at least one base $b$ it excludes convergence to $\frac{b-1}2$.
