The goal of this task given to me is to show that there is no (non-trivial) translation invariant measure on $P(\mathbb{R})$, the power set of $\mathbb{R}$, and I think I almost completed it, but I just can't find a way to prove the very last bit that's missing. But let's start at the beginning.

Let $\mu: P(\mathbb{R}) \to [0, ∞]$ be a measure that satisfies the following conditions:

$(*) \mu([0, 1]) = 1$ and $\mu(x + A) = \mu(A)$ for all $x \in \mathbb{R}, A \subseteq \mathbb{R}$.

The assignment asks me to consider the equivalence relation

$x \sim y :<=> x - y \in \mathbb{Q}$

Out of every equivalence class, we choose a representative $x \in [0, 1]$. Let $X \subseteq [0, 1]$ be the set of representatives that we got that way.

I am now to consider the sets $\mathbb{R} = \cup_{x \in X} (x + \mathbb{Q})$ and $\mathbb{R} = \cup_{q \in \mathbb{Q}} (X + q)$, and find a contradiction by showing that $\mu$ would need to satisfy both $\mu(X) = 0$ aswell as $\mu(X) > 0$.

What I've shown so far: I showed that $\mu(\mathbb{R}) = ∞$ (using the fact that $[0, 1]$ is sent to $1$), aswell as that $\mu(\{a\}) = 0$ for all $a \in \mathbb{R}$. From this, it also follows that all countable subsets of $\mathbb{R}$ are sent to $0$ by $\mu$ (so especially $\mu(\mathbb{Q}) = 0$).

I've also shown that, since $\mu(\cup_{q \in \mathbb{Q}} (X + q)) = ∞$ and since $\cup_{q \in \mathbb{Q}} (X + q)$ is a countable, disjoint union, we have that $\mu(X) > 0$.

The only part I'm still missing is to show that $\mu(X) = 0$ via the fact that $\mathbb{R} = \cup_{x \in X} (x + \mathbb{Q})$, as written above. I just can't get my head around how I could show this. $\cup_{x \in X} (x + \mathbb{Q})$ is a disjoint, but uncountable union because $X$ contains uncountably many elements; therefore, we can't use the $\sigma$-additivity of a measure. I see that $\mu(x + \mathbb{Q}) = 0$ for each $x \in X$, because $\mathbb{Q})$ is countable, but I don't know how that helps me.

I've also thought about decomposing $[0,1]$ into a disjoint union of sets, but that didn't lead anywhere so far. If we "remove" countably many points of $[0,1]$, the remaining set would still be sent to $1$ by $\mu$, for the reasons given above. So how could I conclude that $X$ must be sent to $0$? I'm out of ideas.

  • 1
    $\begingroup$ As far as i can tell the fact you are trying to prove is(maybe a bit more) is in theorem 2.22 of rudin's real and complex analysis, chapter 2. $\endgroup$ – random123 Nov 25 '15 at 9:42
  • $\begingroup$ @random123 Thanks, that helped me solve it indeed. While being slightly different from what I was trying tro prove (the theorem for example used the Lebesgue measure while I was just assuming there is $any$ measure, which made 1 - 2 things a little more difficult; also I had to replace the compact set by a bounded one), the proof itself could be used almost identically. $\endgroup$ – moran Nov 26 '15 at 12:06
  • $\begingroup$ I admit that i did not look carefully into the theorem. I just remembered seeing something similar and suggested the same. $\endgroup$ – random123 Nov 26 '15 at 13:12

One way of doing this is in analogy with the construction of the Vitali set.

Suppose $X$ is defined as in your question. Construct $\tilde{X}$ by taking each $x\in X$, and then translating it by a rational to the interval $[0,\epsilon)$ for some fixed $0<\epsilon<1$. Then the set $\tilde{X}$ is just $X$ where all representatives of the equivalence classes are chosen to lie in $[0,\epsilon)$. Arguing as you did, we can show $\mu(\tilde{X})>0$.

Now consider the set $$E=\cup_{q\in\mathbb{Q}\cap[0,1-\epsilon]}(\tilde{X}+q).$$ Then $E\subset [0,1]$, and so $\mu(E)\le 1$. On the other hand, as the sets $\tilde{X}+q$ are disjoint:

$$\mu(E)=\sum_{q\in \mathbb{Q}\cap[0,1-\epsilon]}\mu(\tilde{X}+q).$$ Applying translational invariance $$=\sum_{q\in \mathbb{Q}\cap[0,1-\epsilon]}\mu(\tilde{X}).$$ The sum is bounded above by $1$, and has countably infinitely many terms of size $\mu(\tilde{X})$. We must therefore have $\mu(\tilde{X})=0$, which contradicts our earlier result that $\mu(\tilde{X})>0$.


I think that the question is not correct. If for each $Y \subset R$ we set $\mu(Y)=+\infty$ if $card(Y)>\aleph_0$ and $\mu(Y)=0,$ otherwise, then $\mu$ stands an example of nontrivial translation invariant measure in $R$ which vanishes on singletons.

If you require that in addition $\mu$ must be $\sigma$-finite, then such stated question is very old and it is not solvable within the theory $ZF$.

For example, if we accept Axiom of Choice then by using Ulam's well known theorem, asserted that on the powerset of the $\aleph_1$ there does not exist a $\sigma$-finite measure which vanishes on singletons, we get a negative answer.

But if we accept Steinhauss-Mycielski determinateness axiom about the existence of winning strategy then following Mycielski and Swierczkowskievery celebrated century theorem asserted that under that axiom each subset of the real axis is Lebesgue measurable we can answer positively on your question(cf. [Mycielski J., Swierczkowski S., On the Lebesgue measurability and the axiom of determinateness, Fund.,54,(1964),67-71]).

  • $\begingroup$ Let take an additive group $G_1$ of $R$ of the cardinality $\aleph_1$, consider equivalent classes, consider any selector . Then by the union of $\aleph_1$ shifts of this selector you cover axis. Your measure must be zero on these shifts by the invariance. Now You can construct a probability measure which is defined on the powerset of $G_1$ and vanishes on singletons. This contradicts to Unlam,s result. $\endgroup$ – George Nov 26 '15 at 21:16
  • $\begingroup$ @moran The set $X$ is Vitali set. This is not Lebesgue measurable but there exists a translation invariant extension $\mu$ of the lebesgue measure for which $\mu(X)=0$. Hence this way is not clever to show the validity of your fact. $\endgroup$ – George Nov 28 '15 at 12:00

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