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I am looking for a non measurable set. I found a set I am almost sure it is not measurable, but am unable to prove it. First we define $f:[0,1]\to [0,9]$ by $$f(0.a_1a_2a_3...)=\begin{cases}\lim_{n\to\infty}(1/n\sum_{i=1}^n a_i) &\text{if the limit exists } \\0 &\text{if the limit does not exist}\end{cases}$$ That means $f$ gives the mean value of the decimals, for example $f(1/3)=3$ and $ f(1/7)=4.5$ Remark: if a number has 2 decimal representations ($0.2$ and $0.1999...$) we define $f(x)=0$ to avoid problems.

Now we define $$A=\{x\in[0,1]:f(x)=4.5\}$$ I am convinced that both $A$ and $A^c$ have an outer measure of $1$, and thus are not measurable, but I can't prove it. Here is what makes me believe their exterior measure is $1$:

For $A$:

-$A$ is uncountable (necessary for a non zero measure)

-$A$ is dense (necessary for an outer measure of $1$) proof: we can take the first n decimals as we want, and make the limit converge to 4.5 by alternating 4's and 5's (for example to find a number in $A$ between 0.2 and 0.201 we can take 0.20454545...)

-We can chose half of the decimals (that means $a_1,a_3,...$) of a number as we wish, and then chose the others to make sure the number is in $A$. This implies $A$ is uncountable, but is much stronger. Proof: simply make $a_{2k}+a_{2k+1}=9$

-If we take a number randomly (that means by choosing $a_1,a_2,...$ randomly), we have a probability of $1$ that this number is in $A$, which means "almost all the numbers" are in $A$.

For $A^c$:

-$A^c$ is uncountable

-$A^c$ is dense. Proof: the numbers with a finite decimal representation are dense in $[0,1]$ and not in $A$

-We can chose any non 1 proportion of the decimals as we wish and still make sure the number is not in $A$. For example, we can set all the decimals $a_k$ such that $k\neq 0$ mod n, and change only the decimals $a_k$ such that $k=0$ mod n, and make sure the number is not in $A$. This means you can chose "almost all the decimals" as you wish, and $A^c$ must be "very big". Proof: set all decimals you are allowed to modify to $0$. If the number is not in $A$, you are done. If the number is in $A$, set all the decimals you can modify to $1$. As this proportion was not $0$, you have increased $f(x)$, thus it is no longer equal to $4.5$, and the number is not in $A$.

That are all properties of $A$ and $A^c$ that may be important for the outer measure I was able to find. For approximations of the outer measure, I only have the trivial properties that they are at most $1$, and their sum is at least $1$.

Can anyone prove or disprove the fact $A$ is measurable? Or can someone find a better upper/lower bound for the outer measures of $A$ and $A^c$?

Thank you for your help, and if you don't understand my notations or claims, feel free to ask, I know it may not all be very clear.

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    If you want a non-measurable set, then you have to invoke Choice somewhere, since it's consistent with ZF that every set of reals is measurable. – 2017-02-26
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    I know I have to invoke Choice somewhere, but I have no idea where or how. – 2017-02-26
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    It would have to be in the *definition* of the set, not in the proof of non-measurability. – 2017-02-26
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    Nitpick: You function is not well defined, since it would yield $f(0.2)=0$, but $f(0.1999\cdots)=9$. This is easily remedied, though, and has no bearing on the problem. – 2017-02-26
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    My definition does not need Choice. Does it mean my set is mesurable? Or can this problem be undecidable? If it is measurable, can someone calculate the outer measures of $A$ and $A^c$? – 2017-02-26
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    @oak no this does not mean it is measurable. You can define plenty of sets whose existence don't need choice but still they might be non measurable. For example define X to be the least non measurable set if V=L and else take the empty set. – 2017-02-26
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    To prove that a definition yields a non measurable set you will more than ZF, for example choice. – 2017-02-26

1 Answers 1

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The set $A$ that you describe is measurable with measure $1$. This follows from the following considerations:

  1. One way to choose a number uniformly at random from the interval $[0,1]$ is to choose each of its digits randomly from the set $\{0,1,\ldots,9\}$. That is, the function $$ f\colon \{0,9\}^\omega \to [0,1] $$ that maps each sequence $(a_1,a_2,\ldots)$ to the number $0.a_1a_2\ldots$ is measure-preserving.

  2. For a randomly chosen sequence from the set $\{0,1,\ldots,9\}$, the sequence of sample means converges to $4.5$ with a probability of $1$. This is a special case of the strong law of large numbers.

Thus $f^{-1}(A)$ has measure $1$ in $\{0,1,\ldots,9\}^\omega$ by statement (2), which proves that $A$ has measure $1$ in $[0,1]$ by statement (1).

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    This implies that the outer measure of $A^c$ is $0$. Is there an explicit construction of a sequence of sets of intervals such that $A^c$ is contained in the intervals, and the sum of the lengths tends to 0? Or can we only prove the existence of such a set because $A$ is measurable? – 2017-02-27