Suppose that $(X,\mathcal{E},\mu)$ is a non-atomic finite measure space (i.e. for every $E \in \mathcal{E}$ with $\mu(E)>0$ there exists $F \subset E$ measurable such that $0<\mu(F) <\mu(E)$.)

a) Prove that for every $ \varepsilon >0$ there is a finite partition of $X$ in measurable subsets $X_1,..,X_n$ such that $\mu(X_i)\leq \varepsilon$.

b) Prove that for every $\alpha \in [0,\mu(X)]$ there exists $E \in \mathcal{E}$ with $\mu(E)=\alpha$.

I guess that a) is given to prove b) more easily. I have the following idea of solution for a) (inspired from this Wikipedia post, which proves b) )

Denote $$ \Gamma = \{ (X_1,..,X_n) : n >> \mu(X)/\varepsilon, X_i \text{ are disjoint }, \mu(X_i)\leq \varepsilon \}$$ ordered by componentwise inclusion. Totally ordered parts $(Y_\alpha)$ of $\Gamma$ have an upper bound element $(\bigcup Y_\alpha^i)_{i=1}^n$ which is still in $\Gamma$.

By Zorn's lemma $\Gamma$ has maximal elements. If a maximal $(X_1..X_n)$ element is not a partition, then we can replace it with something better, of the form $(X_1,..,X_n\cup A)$ where $A \subset X \setminus (X_1 \cup..\cup X_n)$ and $\mu(A)>0$.

My questions are:

1) Is my solution of a) correct? I feel that the part with the upper bound element may not work, since there might be noncountable unions.

2) What is a simpler solution of a)?


7 Answers 7


Clearly b) implies a), so I directly prove b). Let $E\in\mathcal E$ be such that $\mu(E)>0$. Let $A_1\in\mathcal E$ be such that $A_1\subseteq E$ and $0<\mu(A_1)<\mu(E)$. Then either $\mu(A_1)\leq\mu(E)/2$ or $\mu(E\setminus A_1)\leq\mu(E)/2$, so if necessary we consider $E\setminus A_1$ instead of $A_1$, and we can assume that $0<\mu(A_1)<\mu(E)/2$. Now repeat the process with $A_1$ instead of $E$, obtaining a subset $A_2$ of $A_1$ such that $0<\mu(A_2)<\mu(E)/2^2$. Continuing in this way we see that $E$ contains subsets with arbitrarily small positive measure.

Let $\epsilon\in\bigl(0,\mu(X)\bigr)$, and let $A_1\subseteq X$ be such that $0<\mu(A_1)<\epsilon$.

Edit: I expanded this part of the argument, at request of @Michael Greinecker for further clarification.

Suppose that pairwise disjoint subsets $A_1,\dots,A_n$ of $X$ have been chosen such that $D_n=A_1\cup\cdots\cup A_n$ satisfies $\mu(D_n)<\epsilon$. The first paragraph shows that the family

$$\mathcal F_n=\bigl\{C\subseteq X\setminus D_n: 0<\mu(C)<\epsilon-\mu(D_n)\bigr\}$$

is nonempty; however, we cannot determine at this point if the family

$$\mathcal G_n=\bigl\{C\subseteq X\setminus D_n: \frac1n\leq\mu(C)<\epsilon-\mu(D_n)\bigr\}$$

is nonempty as well. We take $A_{n+1}\in\mathcal G_n$ if $\mathcal G_n\ne\emptyset$ and $A_{n+1}\in\mathcal F_n$ otherwise.

If $A=\cup_{n=1}^\infty A_n$, then $\mu(A)=\sum_{n=1}^\infty\mu(A_n)\leq\epsilon$. Suppose that the strict inequality holds, that is $\mu(A)<\epsilon$, and let $B\subseteq X\setminus A$ be such that $0<\mu(B)<\epsilon-\mu(A)$. Then for all $n$ we have $B\subseteq X\setminus D_n$, and since $\mu(D_n)\leq\mu(A)$, it follows that $0<\mu(B)<\epsilon-\mu(D_n)$.

If $N$ satisfies $\frac1N\leq\mu(B)$ then for all $n\geq N$ we have $B\in\mathcal G_n$. Thus, $\mathcal G_n\ne\emptyset$, so by construction necessarily we have $\mu(A_n)\geq\frac1n$ for all $n\geq N$, which is absurd because in this case $\mu(A)\geq\sum_{n=N}^\infty\mu(A_n)\geq\sum_{n=N}^\infty\frac1n=\infty$. This contradiction shows that $\mu(A)=\epsilon$, as desired.

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    $\begingroup$ Could you please report here, a prove from (b) implies (a). $\endgroup$ Commented Jan 21, 2014 at 12:58
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    $\begingroup$ @ZirkoTammado Sketch: given $\epsilon>0$, if $\mu(X)\leq\epsilon$ we are done; otherwise there is a subset $A_1$ with $\mu(A_1)=\epsilon$. Now repeat the process with $X\setminus A_1$ instead of $X$, and so on. The process must terminate after a finite number of steps because $\mu(X)<\infty$. $\endgroup$ Commented Jan 21, 2014 at 13:32
  • $\begingroup$ What does the "whenever possible" mean? And how do you show it is possible often enough without having essentially proven a)? $\endgroup$ Commented Jan 30, 2014 at 8:45
  • $\begingroup$ @MichaelGreinecker See my edit. $\endgroup$ Commented Jan 30, 2014 at 23:50
  • $\begingroup$ Thank you, I got it now. I would however change $D$ to $D_n$ to clarify the dependence on the index. $\endgroup$ Commented Jan 31, 2014 at 0:30

Here is an argument: WLOG $\mu(X) = 1$. Note that it is enough to show that every set can be divided into two pieces of equal measure since then you can get every measure in $[0, 1]$ using countable sums of numbers of the form $1/2^n$. First we show that for any $Y \subset X$ and any $\epsilon > 0$, there is a subset of $Y$ of measure less than $\epsilon$. This is done by dividing $Y$ into two pieces and then dividing the smaller piece into two pieces and so on. In some finite number of steps we will reduce the measure by a large enough factor so eventually we get a piece of measure less than $\epsilon$. Now fix $Y \subset X$. By induction on ordinals $\alpha$, build a transfinite sequence of increasing subsets $Y_{\alpha} \subset Y$ as follows: $Y_0$ is a subset of $Y$ of measure less than $1/2$. At limit stages take union and stop if the union has exactly one half the measure of $Y$. At successor stage $\alpha + 1$, let $Y_{\alpha + 1}$ be $Y_{\alpha}$ union a piece of $Y \backslash Y_{\alpha}$ of measure less than $1/2 - \mu(Y_{\alpha})$. The construction stops in countably many steps because $\mu(Y_{\alpha})$ increases with $\alpha$.

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    $\begingroup$ The indices $\alpha$ are indexing which set ? $\endgroup$ Commented Feb 8, 2015 at 0:29

Here is a quick and dirty nonconstructive proof: Order measurable subsets of $E$ such that $F<G$ iff $F\subseteq G$ and $\mu(F)<\mu(G)$. Take a maximal chain $\mathcal{C}$ in the order. Take any $r\in (0,\mu(E))$. Let $r_1$ be the supremum of values of $\mu$ in $\mathcal{C}$ smaller than $r$ and let $r_2$ be the infimum of values of $\mu$ in $\mathcal{C}$ larger than $r$. Let $(A_n)$ be a sequence in $\mathcal{C}$ with measure increasing to $r_1$ and $(B_n)$ a sequence in $\mathcal{C}$ with measure decreasing to $r_2$. If there is no $R_1\in\mathcal{C}$ with $\mu(R_1)=r_1$, one could add such $R_1$ as $$R_1=\bigcup_n \bigg(A_n\cap\bigcap_m B_m\bigg),$$ contradicting the maximality of $\mathcal{C}$. Similar, we can find $R_2\in\mathcal{C}$ with $\mu(R_2)=r_2$. We are done if $r_1=r_2$, so suppose this isn't the case. By nonatomicity, there is measurable $A\subseteq R_2\backslash R_1$ with $\mu(A)>0$ but then we have either $\mu(R_1\cup A)=r$, $r_1<\mu(R_1\cup A)<r$, or $r<\mu(R_1\cup A)<r_2$, and the last two cases lead to a contradiction.

  • $\begingroup$ Nice simple argument. Note to self: the existence of $\mathcal{C}$ is the Hausdorff maximal principle. $\endgroup$
    – PatrickR
    Commented Aug 17, 2019 at 1:22
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    $\begingroup$ Because $\mathcal{C}$ is a chain, each $A_n$ is contained in each $B_m$, so I think we can just simplify $R_1$ to be the union of the $A_n$. $\endgroup$
    – PatrickR
    Commented Aug 17, 2019 at 1:24

I think first it would help to establish the existance of $\it any$ set $A$ with $0<\mu(A)\leq\epsilon$. This is straightforward. Take any subset $A_1$ of $X$ with $\mu(A_1)>0$. By your definitions such a set must exist. Notice that $A_1^c$ is in your sigma algebra and $\mu(A^1_c)>0$. Then, find a subset $A_2$ of $A_1$ with $0<\mu(A_2)<\mu(A_1)$. Notice that $A_2^c\cap A_1$ is in your sigma algebra and has positive measure. Now do the same for the set $A_1^c$. This process invariably builds up an increasingly refined partition of your space. Since the size of the partition doubles at each step, this means that sooner or later, we must have some partition element with measure less than $\epsilon$. That is, sooner or later $\epsilon 2^n>\mu(X)$. Notice this is where we need the finiteness condition.

Now consider the following process. Start with $X$ and do the above until you find an $X_1$ with $\mu(X_1)\leq\epsilon$. As we said above, you can do it in finitely many steps. Now take $X-X_1$ and repeat the above to find an $X_2$ with $\mu(X_2)\leq\epsilon$. Then look at $X-X_1-X_2$. Ultimately you will get a sequence $X_1,X_2,\ldots$ of your space with $\mu(X_i)\leq\epsilon$. The issue of course is if it becomes a partition as what can happen is the measure of each $X_i$ can decrease too fast, giving a sequence that's NOT a partition. This is fixed below.

We will show that for every $A$ with $0<\mu(A)$, we can find a set $B$ with $B\subset A$ and $\mu(A)/2\leq \mu(B)<\mu(A)$. We know that we can find a $B\subset A$ so that $0<\mu(B)<\mu(A)$. Take $B^c\cap A$, which is again a subset of $A$. Clearly $B\cup (B^c\cap A)=A$ and both are disjoint, so one of them must have measure at least $\mu(A)/2$.

  • $\begingroup$ I was thinking along similar lines, but don't think this works. You get the element smaller than $\epsilon$ using the finite intersection property and non-atomicity (look at nested decreasing sequence, and derive a contradiction). Maybe your argument works too, but that's how I saw this. In the second step though the measures of the sets you define could form an infinite sequence converging to zero; in other words, it could be a countable (not finite) collection. $\endgroup$ Commented Dec 9, 2012 at 19:11
  • $\begingroup$ I believe we need to use that if $A\subsetB$ and $\mu(A) < \mu(B)$, then it is easy to show that there exists $C$, $A\subsetC\subsetB$ such that $\mu(A)<\mu(C)<\mu(B)$, which should lead to finiteness with some more arguments. $\endgroup$ Commented Dec 9, 2012 at 19:12
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    $\begingroup$ I have tried this. The "slightly adjust the above" seems the essential part to me. You need to guarantee that the sets in the partition are not too big and not too small for this method to work. I haven't managed to prove that. $\endgroup$ Commented Dec 9, 2012 at 19:36
  • $\begingroup$ Agreed. I think using my second remark to show (by some contradiction) that question b holds, and use it to show a, is probably how I'd try to tackle this. But moving on personally. $\endgroup$ Commented Dec 9, 2012 at 20:02
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    $\begingroup$ How does the final paragraph fix the problem that $\mu(X_i)$ decreases too fast? $\endgroup$
    – ZQ Wan
    Commented Jan 6, 2017 at 14:57

Proof of (b): Let $$\mathcal P=\{E \in\mathcal E: \mu(E)\le\alpha\}$$ This is a partially ordered set, ordered by "almost-everywhere" inclusion (meaning that $E\le F$ when $\mu(E\setminus F)=0$).

I claim that every chain $\mathcal C \subset\mathcal P$ has an upper bound. Namely, given such a chain, let $\beta=\sup_{E\in \mathcal C}\mu(E)$, and let $E_{1},E_{2},\dots$ be a countable subset of the chain where $\mu(E_{n})\uparrow \beta$. Then $E=\bigcup_{n=1}^\infty E_{n}$ will be the desired upper bound. To see this, given any $F\in\mathcal C$, there are two cases. If for some $n$, we have $E_n\ge F$, then clearly $E\ge F$. Otherwise, $E_n\le F$ for all $n$, meaning $\mu(F)=\beta$ and $$ \mu(F\setminus E)=\lim_n \mu(F\setminus E_n)=\lim_n \mu(F)-\mu(E_n)=\beta-\beta=0. $$ so again, $F\le E$.

We have shown every chain has an upper bound, which by Zorn's Lemma proves that $\mathcal P$ has an upper bound, $G$. We can then show that $\mu(G)=\alpha$. If not, then using the fact that $\mu$ is nonatomic* we could find a subset $H$ of $X\setminus G$ where $0<\mu(H)<\alpha-\mu(G)$, so that $G\cup H\in \mathcal P$, contradicting the maximality of $G$.

*Here we use the lemma that if $\mu$ is nonatomic and $E\in \mathcal E$, then there exist subsets of $E$ of arbitrarily small positive size. This can be easily proven from the definition of non-atomic.

  • $\begingroup$ This is old but anyway you can not ensure that the sequence of sets is increasing. $\endgroup$
    – EQJ
    Commented Oct 26, 2017 at 1:52
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    $\begingroup$ @YTS Yes you can. The sets $E_1,E_2,\dots$ are all elements of the chain, $\mathcal C$. By definition of a chain, for every $i\le j$, we either have $E_i\le E_j$ or $E_i\ge E_j$. Since $\mu(E_i)$ is increasing (by construction), this means $E_i\le E_j$. $\endgroup$ Commented Oct 26, 2017 at 17:03

I have found a very nice proof of this theorem in this Reddit comment. Again, I will take it as given that any subset of $X$ has a further subset with arbitrarily small measure. To prove the main result, we first show:

Lemma: For any $\alpha\in [0,\mu(X)]$, there exists a set $A$ for which $\frac12\alpha\le \mu(A)\le \alpha$.

Proof: Let $\def \C{\mathcal{C}}\C$ be the collection of subsets $A$ of $X$ for which $\mu(A)< \frac12\alpha$. If $\C$ is not closed under unions, then the lemma is proved. To see this, note that if $A,B\in C$, but $A\cup B\notin \C$, then it must be the case that $\frac12\alpha\le \mu(A\cup B)\le \alpha$. Therefore, we assume $\C$ is closed under binary unions. Taking limits, this implies $\C$ is closed under countable unions.

Let $\beta=\sup_{C\in \C}\mu(C)$. There exists a sequence of sets $B_1,B_2,\dots$ for which $\mu(B_n)\nearrow \beta$. Letting $B=\bigcup_n B_n$, this implies $\mu(B)=\beta$, and since $B\in \C$, we have $\beta<\frac12\alpha$. But then we can find a subset $E$ of $X\setminus B$ whose measure is less than $\frac12\alpha-\beta$, which would imply $B\cup E\in \C$, contradicting the fact that $B$ attains $\sup_{C\in \C}\mu(C)$.$\tag*{$\square$}$ With the Lemma under our belt, we can construct a sequence of pairwise disjoint sets $A_1,A_2,A_3\dots$ so that for all $n\in \{1,2,3,\dots\}$, we have $$ \mu(A_n)\in [\tfrac12\alpha_n,\alpha_n],\quad\text{where}\\ \alpha_n=\alpha-(\mu(A_1)+\mu(A_2)+\dots+\mu(A_{n-1})) $$ Then it is easy to show that $\alpha-\mu(A_1\cup\dots\cup A_n)\le (1/2)^n\alpha$, which implies $\mu(\bigcup_{n=0}^\infty A_n)=\alpha$.


Is my solution of a) correct?

Your suspicion that your solution might be incorrect is correct.

What is a simpler solution of a)?

You can use a greedy algorithm in the hope that we finish the job in finitely many steps.

Choose $X_1$ with $\mu(X_1) < \epsilon$ greedily, i.e., first let $J_1$ be the non-empty collection of all $X'_1$ with $0 < \mu(X'_1) < \epsilon$, then choose $X_1 \in J_1$ in such a way that

(1) $\mu(X_1) \ge 0.9 \mu(X'_1)$ for every $X'_1 \in J_1$,

(2) and if $X \in J_1$ then we simply choose $X_1 = X$.

If we are not done yet, i.e. if $X - X_1$ is not a null set, choose $X_2$ inside $X - X_1$ with $\mu(X_2) < \epsilon$ greedily, i.e., first let $J_2$ be the non-empty collection of all $X'_2 \subset X - X_1$ with $0 < \mu(X'_2) < \epsilon$, then choose $X_2 \in J_2$ in such a way that

(1) $\mu(X_2) \ge 0.9 \mu(X'_2)$ for every $X'_2 \in J_2$,

(2) and if $X - X_1 \in J_2$ then we simply choose $X_2 = X - X_1$.

If we are not done yet, i.e., if $X - X_1 - X_2$ is not a null set, we go on to choose $X_3 \subset X - X_1 - X_2$ with $\mu(X_3) < \epsilon$ greedily.

This process finishes in finitely many steps, producing the desired finite partition, because otherwise we end up with either

(1') an infinite sequence of $X_i$ whose union has measure $< 1$ (let's assume $\mu(X) = 1$)


(2') an infinite sequence of $X_i$ whose union has measure 1.

but the possibility (1') is prevented by the condition (1) and (2') by (2).


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