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I know that this question has been discussed in this page(Measure of reals in $[0,1]$ which don't have $4$ in decimal expansion), and I have read this question. I get the solution. However, I thought about another way, and it leads me to another answer, which I am confused. Btw, the question is given as the following:

Let A be the subset of $[0, 1]$ which consists of all numbers which do not have the digit $4$ appearing in their decimal expansion. Find $m(A)$.

The answer is $0$. However, I thought about this solution:

The set of all numbers not containing the digit $4$ in the decimal expansion, are numbers that are only containing $0, 1, 2, 3, 5, 6, 7, 8, 9$ in their decimal expansion. Set a function $f:\mathbb{N}\to\mathbb{N}$ such that: $$f(0)=0, f(1)=1, f(2)=2, f(3)=3, f(5)=4, f(6)=5, f(7)=6, f(8)=7, f(9)=8$$ Then, for any number, say $x=0.a_1a_2a_3a_4, ...$ in the set $A$, let $f(x)=0.f(a_1)f(a_2)f(a_3)f(a_4)..$. Then, this defines a one-to-one map from the set $A$ to $[0, 1]$, as the range of this function can be viewed as a hexadecimal of all numbers in $[0, 1]$. Therefore, the measure of set $A$ and the measure of set $[0, 1]$ must be the same, and $m(A)=1$.

Now, I know that my proof is wrong. Where am I wrong? Is it the 'one-to-one map exists, then the measure must be same' part?

Joshua Woo
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2 Answers2

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Just because there is a bijection from one mesurable set to another you cannot say they have the same measure.For example, the Cantor set has measure $0$ and the real line has measure $\infty$ but they there is bijection from one onto the other since they have the same cardinality.

Kavi Rama Murthy
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Is it the 'one-to-one map exists, then the measure must be same' part?

Yes, that's it. For instance, any interval has the same cardinality, and it is very easy to construct bijections between them, but they do not all have the same measure.

Arthur
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