Since two days ago I have tried solve the following problem but without success.
Let $A\subset I$ be the set of all of numbers without the number three in its decimal expansion (Here $I:=[0,1]$). If $\mu$ is the Lebesgue measure, I want to show that $A$ is a measurable set (respect to Borel's $\sigma$-algebra on $\mathbb{R}$) and I want to calculate its measure.
I think that I can to show that $A$ is a measurable set. Consider $B_n:=\{x\in I: x_n=3\}$ where $x_n$ denotes the $n^{th}$-element in the decimal expansion of $x$. So
$$ B_1=[0.3 \ ,\ 0.4)\\ B_2=\bigcup_{i\in10}[0.i3 \ , \ 0.i4)\\ \vdots\\ B_n=\bigcup_{i\in10^{n-1}}[0.i3 \ , \ 0.i4)\\ \vdots $$
It is inmediate to see that $B_n$ is a measurable set. More over $A=\bigcap_{n\ge 1}B_n^c$, so $A$ is measurable set too. Again, it's easy to see that $\mu(B_n)=0.1$ for each $n$ but I get lost when I try to calculate the measure of $A$.
Can someone help me?
P.D: Also, to me, the way that I try to get the solution seems so complicated. Have you another idea to solve the problem?
Measurability: Regarding your argument that the set is measurable, I think that you have a right idea, but your notation is confusing. What does it mean for $i$ to be in $10^{n-1}$? In analysis (rather than naive set theory), $10^{n-1}$ is a number, not a set. I believe that you are trying to get at the idea that $i$ represents the first $n-1$ digits of a number, followed by a $3$. Therefore, I might write your sets as $$ B_n := \bigcup_{j=0}^{10^{n-1}-1} \left[ \frac{10j + 3}{10^n}, \frac{10j+4}{10^n} \right). $$ Otherwise, your argument seems sound: $$ A = \bigcap_{n=1}^{\infty} B_n $$ is the intersection of a countable number of sets, each of which is the finite union of half-open intervals, which are all measurable. Looks good!
The Measure: Unfortunately, while you have shown that $A$ is measurable, your decomposition seems not very helpful for determining the measure—one would probably like to use continuity from above, but your sets overlap too much, hence determining $$ \mu\left( \bigcap_{n=1}^{N} B_n \right) $$ is rather difficult. I would suggest that you attempt a decomposition which is more disjoint. A possible approach is as follows:
In the first step, you remove the interval $[0.3, 0.4)$, which has measure $10^{-1}$. Let $C_1$ denote the complement of this set, i.e. $$ C_1 := \left[ 0, \frac{3}{10} \right) \cup \left[ \frac{4}{10}, 1\right]. $$ Next, remove every number which has a $3$ in the second decimal place. This removes several intervals of length $1/100$. Specifically, there are $9$ intervals to remove (you might think that there are ten such intervals, but you already killed off one in the first step). Define $C_2$ to be the complement of the removed stuff: \begin{align} C_2 &:= [0,0.03) \cup [0.04,0.13) \cup [0.14,0.23) \cup [0.24,0.3) \\ &\qquad\qquad\quad \cup [0.4,0.43) \cup [0.44,0.53) \cup [0.54,0.63) \cup [0.64, 0.73) \\ &\qquad\qquad\quad \cup [0.74, 0.83) \cup [0.84,0.93) \cup [0.94, 1]. \end{align} Again, the key is the observation that $C_2$ is obtained by removing $9$ intervals of length $1/100$ from $C_1$. Thus $$ \mu(C_2) = \mu(C_1) - 9 \cdot \frac{1}{100} = \frac{9}{10} - \frac{9}{100} = \frac{81}{100} = \frac{9^2}{10^2}. $$ Continuing in this manner, we remove the numbers with a $3$ in their hundreds place. This kills off a tenth of what is remaining (think about that for a minute—it should eventually come clear). Letting $C_3$ denote this next set, we get $$ \mu(C_3) = \mu(C_2) - \frac{1}{10} \mu(C_2) = \frac{81}{100} - \frac{81}{1000} = \frac{729}{1000} = \frac{9^3}{10^3}. $$ There is a pattern here: $$ \mu(C_n) = \frac{9^n}{10^n} $$ (this should be proved, but it holds). We then get $$ A = \bigcup_{n=1}^{\infty} C_n, $$ since $C_n$ is a set of numbers which has no $3$s in their decimal expansions before the $n$-th decimal digit. Then, using continuity from above, we get $$ \mu(A) = \lim_{n\to\infty} C_n = \lim_{n\to\infty} \frac{9^n}{10^n} = 0. $$
Addendum: Because this strikes a little close to my area of interest (fractal geometry), the idea above basically comes down to the observation that $A$ is a (generalized) Cantor set. There are actually a great number of tools which make the analysis here a little easier. For each $$j\in\{0,1,2,4,5,6,7,8,9\}$$ (note the lack of $3$), define the contracting similitude $\varphi_j$ by $$\varphi_j(x) = \frac{1}{10}x + \frac{j}{10}. $$ Then define the mapping $\Phi: [0,1] \to [0,1]$ by $$ \Phi(E) = \bigcup_{j} \varphi_j(E). $$ While it may not be obvious, it turns out that the set we are interested is, essentially, the unique compact set $A$ with the property that $$ \Phi(A) = A.$$ (there is a small issue with a countable number of endpoints, but countable implies measure zero, so... meh). Now... notice that if $E$ is a set having the property that $$ \varphi_j(E) \cap \varphi_k(E)$$ is a nullset for $j\ne k$, then $$ \mu(\Phi(E)) = \frac{9}{10} \mu(E). $$ But, after the application of some heavy theorems (the Banach fixed point theorem, in particular), we get $$ A = \lim_{n\to\infty} \Phi^n(E), $$ where $\Phi^n$ denotes the $n$-fold composition of $\Phi$ with itself. Hence $$ \mu(A) = \mu\left( \lim_{n\to\infty} \Phi^n([0,1]) \right) = \lim_{n\to\infty} \mu(\Phi^n([0,1]) = \lim_{n\to\infty} \left( \frac{9}{10} \right)^n \mu([0,1] = 0.$$ Note that there is a sneaky application of some convergence theorem here, which lets us exchange the limit with the measure (either the monotone or dominated convergence theorem could apply, depending on how you want to justify things).