Find out the outer measure of E and also show that it is measurable.

Question

# Let $ E $ be the set of all real numbers in $ (0,1)$ such that each $ x $ has decimal representations with digits a $ 2 $ or a $ 4$ . Find out the outer measure of the set $ E $ and show that $ E $ is measurable set.

Answer 1

Hint, consider the $2^n$ decimal sequences of length $n$ consisting only of $2$ and $4$.

your number must be in one of the intervals of the form $[0.a_1a_2a_3\dots a_n, 0.a_1a_2\dots a_n +\frac{1}{10^n}]$, each such interval has length $10^n$ and there are $2^n$ such intervals (one for each sequence).

Let $A_n$ be the union of all those intervals, then clearly $\mu(A_n)\leq \frac{2^n}{10^n}$. (this set is closed so clearly measurable)

We conclude that $\mu (\bigcap\limits_{n=1}^\infty B_n) = 0$ by convergence of measures.

Since this intersection contains our set (in fact the two sets are equal but we don't need to prove this explicitly) and the lebesgue measure is complete we can conclude that our set is measurable with measure $0$.

Answer 2

Note that $E = {1 \over 10} E + \{{2 \over 10}, {4 \over 10} \}$.

Hence $m^* E \le {2 \over 10} m^* E$, from which we see that $m^* E = 0$.

Then, for any set $A$ we have $m^* A \le m^* (A \cap E) + m^* (A \setminus E)$ and since $m^* (A \cap E) \le m^* E = 0$, we see that $m^* A \le m^* (A \setminus E)$ from which it follows that $m^* A = m^* (A \cap E) + m^* (A \setminus E)$, hence $E$ is measureable.