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# 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.

MAS
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2 Answers2

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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 $0$.

Asinomás
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  • You need to use the outer measure. – copper.hat Dec 31 '16 at 07:30
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    @copper.hat but if you show it is measurable and calculate its "normal" measure then there is no need to use the outer measure. In this case the outer measure argument is absorbed by the proof of completeness for the lebesgue measure via caratheodory. – Asinomás Dec 31 '16 at 07:34
  • Well, I know the answer, but presumably the whole point was for the OP to use the definition of outer measure to see how they apply. Invoking Caratheodory extension before the OP understands how the outer measure works first doesn't seem to have much educational value. In fact, in some regards, outer measure is easier to apply, because you don't need to consider any $\sigma$ algebra first. – copper.hat Dec 31 '16 at 07:42
  • I think it's not completely lacking in educational value, but maybe its just past my bedtime. – Asinomás Dec 31 '16 at 07:44
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    Sorry, I didn't mean to imply no educational value, just that it maybe using too big a tool too early. – copper.hat Dec 31 '16 at 07:45
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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.

copper.hat
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