I have the following question:

Show that there exists a finitely additive measure $\mu $ on $P (\mathbb R)$ such that $\mu([0,1]) = 1$ and $\mu(A+x) = \mu(A) $ for all $A \subseteq \mathbb R$ and for all $x \in \mathbb R$.

My first attempt was using the measure: $$\mu(A) = L\left( \{\ \lambda(A\cap(0,k))\}\ _{k=1}^{\infty}\right)$$ Where $\lambda$ is the Lebesgue measure and $L$ is the Banach limit.
I did manage to show that this is a finitely additive measure, and $\mu([0,1]) = 1$, but this measure is not translational invariant I think I'm close to a solution, but I might be missing another element to show the last part of the question.
My first instinct is maybe to use arithmetic means, but I'm not sure about that approach. I would like some insights regarding this question.
Thank you to all who can help!

  • $\begingroup$ This might be a bit of overkill, but you can use the fact that $S^1$ is an abelian and therefore amenable group to get a rotation invariant finitely additive probability measure on $S^1$. Then identifying $S^1$ with $[0, 1)$ you can then sum over each interval of the form $[n, n+1)$ to get the result. $\endgroup$
    – David Gao
    Commented Apr 3 at 0:32
  • $\begingroup$ It sounds a bit too much since group theory is not part of the subject in the context of the question, mostly functional analysis, measure theory and topology. $\endgroup$ Commented Apr 3 at 11:23
  • $\begingroup$ Not sure that can be completely avoided though. The approach you are taking probably wouldn’t work since $\lambda(A \cap (0, k))$ is not going to be defined for general $A \in P(\mathbb{R})$. $\endgroup$
    – David Gao
    Commented Apr 3 at 13:53

2 Answers 2


This is effectively the approach I mentioned in comments, made more explicit to avoid appealing to the general theory of amenable groups.

We first construct a rotation invariant finitely additive probability measure on $\mathbb{R}/\mathbb{Z} \simeq S^1$. Let $\pi: \mathbb{R} \to \mathbb{R}/\mathbb{Z}$ be the canonical quotient map. Fix $\{s_\lambda\} \subset \mathbb{R}$ s.t. $1 \notin \{s_\lambda\}$ and $\{1\} \cup \{s_\lambda\}$ is a $\mathbb{Q}$-basis of $\mathbb{R}$. For each finite set $F = \{r_1, \cdots, r_n\} \subset \mathbb{R}/\mathbb{Z}$ and $k \in \mathbb{N}_+$, we shall define another finite set $G_{F, k} \subset \mathbb{R}/\mathbb{Z}$ as follows: Let $\tilde{F} = \{\tilde{r_1}, \cdots, \tilde{r_n}\}$ be chosen s.t. $\tilde{r_i} \in \mathbb{R}$ satisfies $\pi(\tilde{r_i}) = r_i$ for each $1 \leq i \leq n$. Since $\tilde{F}$ is finite, there exists $\{s_1, \cdots, s_N\}$, finitely many elements from $\{s_\lambda\}$, such that $\tilde{F} \subset \mathrm{span}_\mathbb{Q}\{1, s_1, \cdots, s_N\}$. We may then write $\tilde{r_i} = \frac{q_{i0}}{p_{i0}} + \sum_{j=1}^N \frac{q_{ij}}{p_{ij}} s_j$ where $p_{ij} \in \mathbb{N}_+$ and $q_{ij} \in \mathbb{Z}$ for each $1 \leq i \leq n$. By replacing $p_{ij}$ by their least common multiple, we may assume $p_{ij}$ are all the same, i.e., we write $\tilde{r_i} = \frac{q_{i0}}{p} + \sum_{j=1}^N \frac{q_{ij}}{p} s_j$ where $q_{ij} \in \mathbb{Z}$ an $p \in \mathbb{N}_+$ is fixed and independent of $i$ and $j$. Let $Q = \max\{|q_{ij}|: 1 \leq i \leq n, 1 \leq j \leq N\}$.

We observe that, since $\{1, s_1, \cdots, s_N\}$ is linearly independent over $\mathbb{Q}$, no nontrivial $\mathbb{Q}$-linear combination of $\{s_1, \cdots, s_N\}$ can be a rational number. Thus, we have, for any $l \in \mathbb{N}_+$, $F_l = \pi(\{\frac{a_0}{p} + \sum_{j=1}^N \frac{a_j}{p} s_j: a_0 = [0, p) \cap \mathbb{Z}; a_1, \cdots, a_N \in [-l, l] \cap \mathbb{Z}\})$ has cardinality $|F_l| = p(2l+1)^N$. Thus, as $l \to \infty$, $\frac{|F_{l+Q}|}{|F_l|} \searrow 1$. Choose $l$ large enough s.t. $\frac{|F_{l+Q}|}{|F_l|} < 1 + \frac{1}{2k}$ and set $G_{F, k} = F_l$.

Note that, for any $r \in F$, by definition $r + G_{F, k} \subset F_{l+Q}$, so $|(r+G_{F,k}) \triangle G_{F,k}|/|G_{F,k}| < \frac{1}{k}$. Let $P_f(\mathbb{R}/\mathbb{Z})$ be the collection of finite subsets of $\mathbb{R}/\mathbb{Z}$, and for any $F \in P_f(\mathbb{R}/\mathbb{Z})$, let $L_F \subset P_f(\mathbb{R}/\mathbb{Z})$ be the collection of elements of $P_f(\mathbb{R}/\mathbb{Z})$ which contains $F$. Then $\{L_F: F \in P_f(\mathbb{R}/\mathbb{Z})\}$ has the finite intersection property (i.e., any finite intersection of $L_F$ is nonempty), so we may extend it to an ultrafilter $\mathcal{U}$ on $P_f(\mathbb{R}/\mathbb{Z})$. Choose any free ultrafilter $\mathcal{V}$ on $\mathbb{N}_+$.

We may now define $\mu_1: P(\mathbb{R}/\mathbb{Z}) \to [0, 1]$ as follows:

$$\mu_1(A) = \lim_{F \to \mathcal{U}} \lim_{k \to \mathcal{V}} \frac{|A \cap G_{F,k}|}{|G_{F,k}|}$$

It is easy to verify that $\mu_1$ is a finitely additive probability measure. I claim that $\mu_1$ is rotation invariant as well. Observe that $L_{\{r\}} \in \mathcal{U}$. Let $r \in \mathbb{R}/\mathbb{Z}$ and $\epsilon > 0$ be arbitrary. Let $F \in L_{\{r\}}$, i.e., $r \in F$. It suffices to prove that, for large enough $k \in \mathbb{N}_+$,

$$|\frac{|A \cap G_{F,k}|}{|G_{F,k}|} - \frac{|(- r + A) \cap G_{F,k}|}{|G_{F,k}|}| < \epsilon$$

I claim that this holds whenever $\frac{1}{k} < \epsilon$. Indeed, since $r \in F$, we have $|(r+G_{F,k}) \triangle G_{F,k}|/|G_{F,k}| < \frac{1}{k} < \epsilon$, so,

$$\begin{split} |\frac{|A \cap G_{F,k}|}{|G_{F,k}|} - \frac{|(- r + A) \cap G_{F,k}|}{|G_{F,k}|}| &= \frac{||A \cap G_{F,k}| - |A \cap (r + G_{F,k})||}{|G_{F,k}|}\\ &\leq \frac{|(A \cap G_{F,k}) \triangle (A \cap (r + G_{F,k}))|}{|G_{F,k}|}\\ &= \frac{|A \cap (G_{F,k} \triangle (r + G_{F,k}))|}{|G_{F,k}|}\\ &< \epsilon \end{split}$$

This proves $\mu_1$ is a rotation invariant finitely additive probability measure on $\mathbb{R}/\mathbb{Z} \simeq S^1$. Clearly, this implies $\mu_1(\{r\}) = 0$ for all $r \in \mathbb{R}/\mathbb{Z}$. Now the result follows easily by defining $\mu: P(\mathbb{R}) \to [0, +\infty]$ as follows,

$$\mu(A) = \sum_{k = -\infty}^\infty \mu_1(\pi(A \cap [k, k+1)))$$

  • $\begingroup$ I believe something like this is basically unavoidable - a translation invariant finitely additive measure $\mu$ on $P(\mathbb{R})$ with $\mu([0, 1]) = 1$ immediately yields on a rotation invariant finitely additive probability measure on $P(S^1)$ by restricting $\mu$ to $[0, 1)$ and identifying $[0, 1)$ with $S^1$ canonically. But the existence of the latter is the definition of $S^1$ being amenable. Basically, the question is the same as asking to prove the amenability of $S^1$. $\endgroup$
    – David Gao
    Commented Apr 3 at 16:26
  • $\begingroup$ Nice, +1. The amenability of $S^1$ also follows directly from the Markov-Kakutani fixed point theorem, as it is a compact abelian group. $\endgroup$ Commented Apr 3 at 17:18
  • $\begingroup$ @JoseAvilez You’re right. Though there’s no need to mention $S^1$ is compact as I’m regarding $S^1$ as a discrete group (which is necessary since I’m trying to define the measure on all subsets of $S^1$, rather than just Borel subsets), but the Markov-Kakutani fixed point theorem would apply regardless. The answer was just written in this way to be as self-contained as possible, without referencing more advanced machinery. $\endgroup$
    – David Gao
    Commented Apr 3 at 19:06
  • $\begingroup$ I feel like this solution is way more cumbersome than it should be since the course I'm taking does not focus on group theory or abstract algebra. Somebody else proposed to me a solution that appears to be simpler, and I'll probably post it tomorrow. $\endgroup$ Commented Apr 16 at 21:00

A colleague of mine proposed the following solution:
Let $V$ be the space of measurable functions $f:\mathbb R\rightarrow \mathbb R$. Define the mapping $T : V \rightarrow \mathbb R$ by: $$T(f) = \int fd\lambda$$ Where $\lambda$ is the Lebesgue measure.
Moreover, we'll define $\Gamma_x :\mathbb R^{\mathbb R} \rightarrow \mathbb R^{\mathbb R}$ by $(\Gamma_xf)(y) = f(x+y)$. We'll also define $G = \{\Gamma_x : x\in \mathbb R \}.$
It can be shown that $T$ can be extended to a linear functional $\tilde{T}: \mathbb R^{\mathbb R} \rightarrow \mathbb R$ s.t. $\tilde{T} |_V = T,$ with $\tilde{T}(\Gamma_xf) = \tilde{T}(f)$ for all $\Gamma_x \in G$ (This is part of a result we've learned in class).
We'll define $\mu : P(\mathbb R)\rightarrow [0,\infty)$ by $\mu(A) = \tilde{T}(\chi_A).$ This is a finite measure because it represents a functional, and from its construction, it fulfills the needs of the question.
There may be some hidden details that I did not expand upon here, since this solution is based on a statement we've learned in class, but I believe the idea is understood.

  • $\begingroup$ This proof doesn’t really work as is. First of all, $T$ is not well-defined, since not all measurable functions are integrable. In fact, it’s not hard to show that any $\mu$ satisfying the conditions you wrote down cannot be finite, so you cannot just represent it as a linear functional on the space of all functions. This could potentially be fixed, by, for example, restricting to $[0, 1)$ and bounded functions. Then just use an infinite sum to extend to all subsets of $\mathbb{R}$. (Like I did in the final paragraph of my answer.) $\endgroup$
    – David Gao
    Commented Apr 16 at 22:43
  • $\begingroup$ Another thing that’s not working as is (even after restricting to $[0, 1)$) is that why should $\mu$, as defined, to be positive. $\tilde{T}(\chi_A)$ could be negative if $\tilde{T}$ is just some arbitrary linear functional extending $T$, after all. I presume the extension result you cited from your class can be used to address this, but this is not just immediate from what you wrote. $\endgroup$
    – David Gao
    Commented Apr 16 at 22:46
  • $\begingroup$ A final comment: the reason $T$ can be extended to $\tilde{T}$ that’s invariant under $G$ in the first place is precisely because $G$ is amenable, which basically means there’s a proof hidden within this fact that’s just the argument I gave, or the argument Jose gave in comments under my answer. Of course, if you already saw this fact in class, you can just directly apply this, but this is a context for your question that you should have mentioned, as otherwise we would have assumed it’s a fact that needs a proof. $\endgroup$
    – David Gao
    Commented Apr 16 at 22:52
  • $\begingroup$ (This is also why this only works in dimensions $1$ and $2$, while in dimension $3$ there’s no finitely additive measure extending the Lebesgue measure that’s invariant under translations and rotations - the group generated by translations and rotations in dimension $3$, the analogue of your $G$, is not amenable in that case.) $\endgroup$
    – David Gao
    Commented Apr 16 at 22:54
  • $\begingroup$ I see, but I think this approach is more in line with the course I'm taking. Your solution, while seems correct, is not in line with what we learned in our course (which is focused mainly on functional analysis). I think this is what our instructor was hinting at, but this may need some improvement. $\endgroup$ Commented Apr 17 at 8:42

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