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Friends

Just came across an exercise that asks to show that set of numbers in $[0,1]$ which possess decimal expansion not containing the digit $5$ has measure zero ?

How do i approach this?

I tried thinking of relating this to Cantor set but, I see that numbers whose decimal expansion contains the digit $5$ are bith contained in the Cantor set as well in its complement too, I thought f it was in the Cantor set and since measure of Cantor set is zero, so it's measure will be zero but this does not happen.

In fact how is this set measurable?

Any idea?

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    any resource where i can practice these type of questions? – 2017-02-21
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    If it's measure is zero it's cardinality is equal to or less than the naturals. – 2017-02-21
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    Uh, plenty of uncountable sets have measure zero. – 2017-02-21
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    @ZelosMalum Have you met my friend, the [Cantor set](https://en.wikipedia.org/wiki/Cantor_set)? (Which happens to be the set of numbers that do not have a $1$ in their ternary expansion.) – 2017-02-21
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    Hint: You can easily find a superset of your no-digit-5 set that has measure 0.9. Maybe you can then find another, smaller superset that has measure $0.9\cdot0.9$, and then continue. – 2017-02-21
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    Not that it matters for the final measure, but is it $0.5$ or $0.49999\ldots$? – 2017-02-21

2 Answers 2

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See, study the complement of the set. That is, look at integers which contain $5$ in their decimal expansion.

The first such category we can think of is: Those that have $5$ as the first digit. This is the set of numbers $[0.5,0.6)$. This has measure $0.1$, so the left over measure is $0.9$.

Now, from the remaining set, remove the set of all numbers with second digit $5$. This consists of $[0.05,0.06), [0.15,0.16) \ldots [0.95,0.96)$ without $[0.55,0.56)$, since that was already removed earlier. Now, each of these has measure $0.01$, so we have removed $0.09$ more from the system. Hence, the left over is $0.81$.

By induction, prove the following : at the $n$th step, the set left over has measure $\frac{9^n}{10^n}$. Now, as $n \to \infty$, we see that the given set has measure zero (I leave you to rigorously show this, you can use the Borel-Cantelli lemma). This also incorporates the fact that the given set is measurable, since it's measure is computable(and is $0$).

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Denote $E_2^c =$ the set of real numbers $x$ on the interval $[0, 1]$ such that in the decimal form of $x = 0.i_1 i_2 i_3 \dots$ there is no digits $2$.

Write $x \in E_2^c$ in decimal form [ x = 0. i_1 i_2 i_3 \dots, \ \ i_k = 0, 1, 3, 4, 5, 6, 7, 8, 9, \ \ x = \sum_{k=1}^{\infty} \frac{i_k}{10^k}, ] and assume that the first digit $2$ appeals at $k$th decimal of $x$.

Denote $E_{2,1}$ as the set of real numbers $x$ on the interval $[0, 1]$ such that the first digit $2$ appeals at $1$th decimal of $x$: $E_{2,1}= \{x | x = 0. 2 \dots\} = [0.2, 0.3)$, then $m(E_{2,1}) = 10^{-1}$.

Denote $E_{2,2}$ as the set of real numbers $x$ on the interval $[0, 1]$ such that the first digit $2$ appeals at $2$th decimal of $x$: $E_{2,2} = \{x | x = 0. i_1 2 \dots\}= \cup_{i_1} [0.i_1 2, 0.i_1 3)$, $i_1 \in \{0, 1, 3, 4, 5, 6, 7, 8, 9\}$, then $m(E_{2,2}) = 9 \times 10^{-2}$;

Inductively, \begin{align*} E_{2,k} &= \{x | x = 0. i_1 \dots i_{k-1} 2 \dots\} \\ &= \cup_{i_1, \dots, i_{k-1}} [0. i_1 \dots i_{k-1} 2, 0. i_1 \dots i_{k-1} 3), \end{align*} where $i_1, \dots, i_{k-1} \in \{0, 1, 3, 4, 5, 6, 7, 8, 9\}$, then $m(E_{2,k}) = 9^{k-1} \times 10^{-k}$;

Define $E_2 =$ the set of real numbers $x$ on the interval $[0, 1]$ such that in the decimal form of $x = 0.i_1 i_2 i_3 \dots$ there is digits $2$.

Then the measure of $E_2$ is [ m(E_2)= m(\cup_{k=1}^{\infty} E_{2,k})= \sum_{k=1}^{\infty} m(E_{2,k})=\sum_{k=1}^{\infty} 9^{k-1} \times 10^{-k}= 1. ]

Thus, the measure of $E_2^c$ is [ m(E_2^c) = m([0, 1] \setminus E_2) = m([0, 1]) - m(E_2)= 1 - 1 = 0. ]