attempt to construct a uncountable measure zero set I am trying to construct a uncountable measure zero set
this is my 1st attempt:
let $A=${$x\in \Bbb R | x $'s decimal places has infinitely many zero's }
for example $1=1.000...\in A$, $1.010101....\in A$ and $\pi \in A$
it seems like to me this set is uncountable, what about measure of $A$
is $m(A)=0$ or $m(A)>0$ ?
edit-
I am trying to construct the set without using cantor set
 A: A cantor set is a zero-set and uncountable and can be built as follows:consider interval $[0,1]\in R$. Equipartition it by 3 subsets ,namely, $[0,1/3],[1/3,2/3],[2/3,1]$ and remove the middle sub-interval [1/3,2/3].
Perform the same procedure on remaining intervals $[0,1/3],[2/3,1]$ and construct intervals $[0,1/9],[2/9,1/3],[2/3,7/9],[8/9,1]$. Following this idea till infinity you will finally have an uncountable zero-set.
A: Hint:
A classic example is Cantor's ternary set:
$$K_3=\biggl\{\sum_{i\ge 1}\frac{x_i}{3^i}\;\Bigm|\forall i\ge 1, x_i\in\{0,2\}\biggr\}$$
i.e. the set of real numbers in $[0,1]$ which a base $3$ expansion without $1$ among the ternary digits.
A: Since your question seems to be about your choice of $A$ in particular, I'll address that instead of supplying a different option (which others have done very well).
The complement of your $A$ consists of all $x$ with only finitely many zeroes. All such $x$ are rational (because, for example, $0.10010111111\ldots$ is the same as $0.10011$). The set of all rationals has measure $0$, so the complement of $A$ has measure zero. So $A$ has (very) positive measure. Others have suggested the Cantor set instead, which would be an excellent option.
