# Constructing an outer measure on a set whose measurable sets are exactly a given sigma algebra on the set.

Consider an arbitrary set $$X$$ and an arbitrary $$\sigma$$-algebra $$\mathcal{M}$$ on $$X$$.

My question is that can one construct an outer measure on the set $$X$$ whose measurable sets is exactly the collection $$\mathcal{M}$$.

I tried to find an answer for finite sets and found this proposition to be true.

The solution is, let $$X$$-finite set, $$\mathcal{M}$$-algebra on $$X$$(and hence a $$\sigma$$-algebra on $$X$$) and $$\mu_{0}(A)=|A|$$ (cardinality of A) $$\forall A \in \mathcal{M}$$. It is easy to verify that $$\mu_{0}$$ is a pre-measure on the algebra $$\mathcal{M}$$ as $$\mu_{0}(\emptyset)=0$$ and it is countably additive (over here only finite additivity suffices).

Thus we construct the outer measure on $$\mathcal{P}(X)$$ using $$\mu_{0}$$, call it $$\mu^*$$.

Let $$B\subset X(\notin \mathcal{M})$$. So $$(X\setminus B) \notin \mathcal{M}$$. Then $$\mu^*(B)>|B|$$ as all the elements of $$\mathcal{M}$$ containing $$B$$ has higher cardinality.

If possible $$B$$ is $$\mu^*$$-measurable.

So we can check $$|X|=\mu^*(X)= \mu^*(B) + \mu^*(X\setminus B)>|B|+|X\setminus B|=|X|$$. Hence this is a contradiction and hence $$B$$ is not $$\mu^*$$-measurable.

Hence the only $$\mu^*$$-measurable sets are the sets in $$\mathcal{M}$$.

I have no idea how to proceed with this problem for infinite sets and maybe more general cases. Any kind of help and idea is highly appreciated.

Thanks.

Edit: I also figured out that even in infinite sets, if the concerned $$\sigma$$- algebra is a finite one, we can define a pre-measure on it in the same way and check that these are the only measurable sets.

• I answered this exact question here: math.stackexchange.com/questions/2839002/… Commented Feb 11, 2020 at 21:12
• @mathworker21 your answer is just short of being complete though. It would be nice to have a counterexample and it would be nice if you proved the necessary part of your classification there Commented Feb 12, 2020 at 0:05

Second attempt for a counter-example. :-)

Let $$X=X_0\times\{1,2\}$$ with some uncountable set $$X_0$$, and let $$\mathcal{M}=\{A\times\{1,2\}:~A\subset X_0,~\text{either A or X_0\setminus A is countable.}\}$$

Suppose that there is an outer measure $$\mu^*$$ on $$P(X)$$ such that $$\mathcal{M}$$ is precisely the set of $$\mu^*$$-measurable sets.

Notice that for every $$x\in X_0$$, the sets $$\{(x,1)\},\{(x,2)\}\notin\mathcal{M}$$, so these sets are not measurable; therefore $$\mu^*\big(\{(x,1)\}\big)>0$$, $$\mu^*\big(\{(x,2)\}\big)>0$$ and $$\mu^*\big(\{(x,1),(x,2)\}\big)>0$$.

Take some $$A_0\subset X_0$$ such that both $$A_0$$ and $$X_0\setminus A_0$$ are uncountable, and let $$A=A_0\times\{1,2\}$$. Obviously $$A\notin\mathcal{M}$$, so $$A$$ is not measurable. Hence, there is some $$B\subset X$$ such that $$\mu^*(B\cap A)+\mu^*(B\setminus A)>\mu^*(B)$$. It follows that $$m:=\mu^*(B)$$ is finite.

Let $$B_0=\{x\in X_0:\text{ (x,1)\in B or (x,2)\in B}\}$$ be the projection of $$B$$ on $$X_0$$. We shall prove that $$B_0$$ is countable.

Take a positive integer $$k$$ and arbitrary elements $$c_1,\ldots,c_n\in B_0$$ such that $$\mu^*\big(B\cap\{(c_i,1),(c_i,2)\}\big)>\frac1k$$. Let $$C=\{c_1,\ldots,c_n\}\times\{1,2\}$$; since $$\{(c_i,1),(c_i,2)\}$$ is measurable, we get $$m=\mu^*(B) \ge \mu^*(B\cap C)=\sum_{i=1}^n \mu^*\big(B\cap\{(c_i,1),(c_i,2)\}\big)>\frac{n}{k},$$ so $$n. Hence, there are only finitely many elements $$c\in B_0$$ such that $$\mu^*\big(B\cap\{(c_i,1),(c_i,2)\}\big)>\frac1k$$.

Since $$\mu^*\big(B\cap\{(c,1),(c,2)\}\big)>0$$ for all $$c\in B_0$$, this proves that $$B_0$$ is countable.

Now we can replace $$A$$ by $$A'=A\cap(B_0\times\{1,2\})\in\mathcal{M}$$. Note that $$B \cap A = B \cap A'$$, so that we get a contradiction by $$\mu^*(B) < \mu^\ast(B\cap A)+\mu^*(B\setminus A) = \mu^\ast (B\cap A')+\mu^*(B\setminus A') = \mu^*(B).$$

• Oops, you are right, case I is wrong. I will delete the answer. Commented Feb 18, 2020 at 11:22
• Now I have an improved, completely re-written answer. :-) Commented Feb 18, 2020 at 14:40
• This is really nice, thanks! One minor comment which might be helpful for other people: What is used when showing that $\mu^\ast(B\cap C) = \sum_i \mu^\ast (B \cap \{(c_i,1),(c_i, 2)\})$ is the following: If $A_1,\dots,A_n$ are disjoint and measurable and $A = \bigcup_i A_i$, then $\mu^\ast (B \cap A) = \sum_i \mu^\ast(B \cap A_i)$ for arbitrary $B$. This can be proved by induction, noting that since $A_{n+1}$ is measurable, $\mu^\ast(B \cap A) = \mu^\ast (B \cap A \cap A_{n+1}) + \mu^\ast ( (B \cap A) \setminus A_{n+1}) = \mu^\ast (B \cap A_{n+1}) + \mu^\ast (B \cap \bigcup_{i=1}^n A_i)$. Commented Feb 18, 2020 at 15:25
• @user141614 I just have a doubt but the construction is fantastic. How can you assume that $\mu^*(B)=m$ is finite? Commented Feb 20, 2020 at 12:11
• $m=\mu^*(B)$ cannot be infinity because it is less than $\mu^*(B\cap A)+\mu^*(B\setminus A)$. Commented Feb 20, 2020 at 13:13

Under certain additional assumptions it is possible to construct an outer measure s.t. the collection of $$\mu^*$$-measurable sets coincides with $$\mathcal{M}$$ (compare your example with the following result). Suppose that $$(X,\mathcal{M})$$ is endowed with a measure $$\mu$$ and consider the usual construction of the corresponding outer measure $$\mu^*$$, i.e. for $$A\subset X$$, \begin{align} \mu^*(A):\!&=\inf\!\left\{\sum_j \mu(B_j):A\subset \bigcup_j B_j, \{B_j\}\subset\mathcal{M}\right\} \\ &=\inf\{\mu(B):A\subset B\in\mathcal{M}\}. \end{align}

If $$(X,\mathcal{M},\mu)$$ is complete and $$\mu$$ is $$\sigma$$-finite, then $$\mathcal{M}$$ coinsides with $$\mathcal{M}^*$$.

Proof. Suppose that $$\mu(X)<\infty$$. For $$A\in \mathcal{M}^*$$, there exists $$B\in \mathcal{M}$$ s.t. $$A\subset B$$ and $$\mu^*(A)=\mu(B)^{(1)}$$. Thus, $$\mu^*(B\setminus A)=0$$ and (using the same argument) there exists $$C\in \mathcal{M}$$ s.t. $$B\setminus A \subset C$$ and $$\mu(C)=\mu^*(B\setminus A)=0$$. Since $$\mu$$ is complete, $$A=B\setminus(B\setminus A)\in \mathcal{M}$$. As for the general case write $$X=\bigcup_j X_j$$ with $$\mu(X_j)<\infty$$. Then, using the above argument, $$\mathcal{M}^*\ni A=\bigcup_j (A\cap X_j)\in\mathcal{M}$$.

This shows that $$\mathcal{M}^{*}\subseteq\mathcal{M}$$. The other inclusion is obvious. $$\square$$

$${}^{(1)}$$ By the definition of $$\mu^*$$ we can find a decreasing family of sets $$\{B_n\}\subset \mathcal{M}$$ s.t. $$A\subset B_n$$ for each $$n$$ and $$\mu^*(A)=\mu^*(B)=\mu(B)$$, where $$B\equiv\bigcap_{n\ge 1}B_n\in\mathcal{M}$$.

• "reach enough"? Commented Feb 11, 2020 at 21:13
• I downvoted (for now). It seems you're assuming there's a measure present. I'm very confused by your answer. You say that the sought-after result is not possible, but then give a proof of sometimes when it is? You should prove it is not always possible if that is what you claim. Maybe I'm just being stupid, but right now I'm basking in confusion Commented Feb 11, 2020 at 21:16
• See the example given by the OP ($\mu_0$).
– user140541
Commented Feb 11, 2020 at 21:22
• I still don't get it. OP is starting with a sigma-algebra on a finite set, then defining $\mu_0$. I'm starting to get irked. I don't know why you get to suppose that $(X,\mathcal{M})$ is endowed with a measure $\mu$. The question is: My question is that can one construct an outer measure on the set 𝑋 whose measurable sets is exactly the collection M. That's it. I don't know why you're supposing stuff. Commented Feb 11, 2020 at 21:33
• No problem. Think about it as an answer conditional on "supposing stuff". It actually was driven by that example.
– user140541
Commented Feb 11, 2020 at 21:44