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On a measure space $(\Omega,\mathcal{A},\mu)$ a completion of a measure is defined as:

$\{A: A_1\subset A\subset A_2$ with $A_1,A_2\in\mathcal{A}$ and $\mu(A_2\backslash A_1)=0\}$

I'm trying to show that this set is equivalent to:

$\{A\cup N:A\in\mathcal{A}$, and $N\subset B\in\mathcal{A}$ for some $B$ with $\mu(B)=0\}$

and

$\{A\triangle N:A\in\mathcal{A}$, and $N\subset B\in\mathcal{A}$ for some $B$ with $\mu(B)=0\}$ where $\triangle$ is the symmetric difference.

To show that the first definition is equivalent to the second, I showed that some $A_a$ in the first set is in the second set (which was pretty easy since $A_a\in\mathcal{A}$ implies that it is in $\mathcal{A}\cup N$. However, I am having trouble going the other direction.

For the third set, I'm having a hard time seeing the picture.

Thanks for your help.

Michael Greinecker
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John Smith
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2 Answers2

2

I call these families $\mathcal{A}_1,\mathcal{A}_2,$ and $\mathcal{A}_3$.

$\mathcal{A}_1\subset\mathcal{A}_2$:

Let $A_1\subset A\subset A_2$ with $\mu(A_2\backslash A_1)=0$. Then $A=A_1\cup N$ where $N=A\cap (A_2\backslash A_1)$. So $A\in\mathcal{A}_2$.

$\mathcal{A}_2\subset\mathcal{A}_3$:

Take $A\cup N$, where $N$ is a subset of a measurable set $B$ of measure $0$. Let $N'=N\backslash A\subset B\backslash A$. Clearly, $\mu(B\backslash A)=0$. Now $$A\triangle N'=(A\backslash N')\cup (N'\backslash A)=A\cup N'=A\cup N\backslash A=A\cup N.$$

$\mathcal{A}_3\subset\mathcal{A}_1$:

Take $A\triangle N$, where $N$ is a subset of a measurable set $B$ of measure $0$. Let $A_1=A\backslash B$ and $A_2=A\cup B$. Now $A_1=A\backslash B\subset A\backslash N\subset A\backslash N\cup N\backslash A=A\triangle N\subset A\cup N\subset A\cup B=A_2$. Now $A_2\backslash A_1=(A\cup B)\backslash (A\backslash B)=B$, so $\mu(A_2\backslash A_1)=0$.

Michael Greinecker
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  • Hi, Thank you so much for your concise proof. I have two questions. First, when you prove, say, $ \mathcal{A}_1 \subset \mathcal{A}_2$, do you change the expression of $\mathcal{A}_2$ in terms of $\mathcal{A}_1$? – JoZ Sep 12 '20 at 09:17
  • I added a light more details in the first one by $ \mathcal{A}_1 \subset \mathcal{A}_2$: let $A_1 \subset A \subset A_2 $ with $A_1, A_2 \in \mathcal{A}, \mu(A_2\backslash A_1)=0$, then $A=A_1 \cup N$, where $N= A \cap (A_2 \backslash A_1) \subset (A_2\backslash A_1)=B$ having $\mu(B)=0$ and $A_1\in \mathcal{A}$. Not sure if I am doing it correctly. – JoZ Sep 12 '20 at 09:17
  • Also, if my understanding is correct., I am a bit confused with your second proof where you showed $N \backslash A \subset B \backslash A , \mu(B \backslash A)=0$. Should't it be enough to show $N' \subset N \subset B$? – JoZ Sep 12 '20 at 09:19
  • @JoZ If I understand the first part of your comment; these are essentially three independent proofs of implications that jointly imply that all conditions are equivalent. I'm to sure what you ask about the second proof; the goal is to show that a set of the form $A\cup N$ can be written as a set of the form $A\Delta N'$. – Michael Greinecker Sep 12 '20 at 11:18
  • Thank you I think I got the point! – JoZ Sep 12 '20 at 13:29
1

Another way of approaching this is to prove that each of these families ($\mathcal A_1$, $\mathcal A_2$ and $\mathcal A_3$) is equal to $\sigma(\mathcal A \cup \mathcal N)$ which is the sigma algebra generated by $\mathcal A \cup \mathcal N$ where $\mathcal N$ is the collection of subsets of null sets in $\mathcal A$. This approach, which can be more work than the direct approach of simply showing $\mathcal A_1 \subset \mathcal A_2 \subset \mathcal A_3 \subset \mathcal A_1$,, has a benefit of making one see motivations for each of the three definitions.

Let's start with $\mathcal A_1$. To show that this is equal to $\sigma(\mathcal A \cup \mathcal N)$, one only needs to show that $\mathcal A_1$ is a sigma algebra containing $\mathcal A \cup \mathcal N$ and that each element of $\mathcal A_1$ can be obtained by applying countable union, countable intersection, and complement several times to some elements in $\mathcal A \cup \mathcal N$.

It is easy to show that each element of $\mathcal A_1$, $\mathcal A_2$ and $\mathcal A_3$ can be obtained by applying countable union, countable intersection, and complement several times to some elements in $\mathcal A \cup \mathcal N$. Therefore, each of $\mathcal A_1$, $\mathcal A_2$ and $\mathcal A_3$ is a subset of $\sigma(\mathcal A \cup \mathcal N)$.

It is also easy to show that each of $\mathcal A_1$, $\mathcal A_2$ and $\mathcal A_3$ contains $\mathcal A \cup \mathcal N$.

Now it only remains to show that they are sigma algebras. For that, read the following thread first:

Completion of a measure space

The question in that thread is concerned with a collection that you can easily see is equal to $\mathcal A_3$ (Equality follows from $A \Delta B = C \iff B \Delta C = A \iff C \Delta A = B$).

Michael Greinecker's answer in that thread contains a proof of the fact that $\mathcal A_2$ is a sigma algebra.

Jisang Yoo's answer in that thread contains a proof of the fact that $\mathcal A_3$ is closed under finite union, but you can easily generalize the proof to show that it is closed under countable union as well.

As for how to prove that $\mathcal A_1$ is closed under countable union, let $(A_i)$ be a sequence of elements in $\mathcal A_1$. For each $i$, there are $A_i'$ and $A_i''$ in $\mathcal A$ that approximate $A_i$ from below and above in the sense that $A_i' \subset A_i \subset A_i''$ and $\mu(A_i'' \setminus A_i') = 0$.

You'd expect that the union $\bigcup_i A_i$ would be approximated by $\bigcup_i A_i'$ and $\bigcup_i A_i''$ in the same sense. Is it true that $\mu(\bigcup_i A_i'' \setminus \bigcup_i A_i') = 0$ holds? Yes because the final total exception set $\bigcup_i A_i'' \setminus \bigcup_i A_i'$ must be covered by the union of the exception sets we started with, namely $A_i'' \setminus A_i'$, but these are null sets.

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Jisang Yoo
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