Studying Measure Theory in University, I came across the following definition for the completion of a measure space: let $(X,\mathcal{E},\mu)$ be a measure space; then the set $$\overline{\mathcal{E}}=\left\lbrace A\subseteq X:\exists B,C\in\mathcal{E}:A\triangle B\subseteq C\wedge\mu(C)=0\right\rbrace$$ is a $\sigma$-algebra and, extending $\mu$ to $\overline{\mu}$ defined on $\overline{\mathcal{E}}$ by letting $\overline{\mu}(A)=\mu(B)$ for any set $B\in\mathcal{E}$ such that $A\triangle B$ is contained in a null set, the measure space $(X,\overline{\mathcal{E}},\overline{\mu})$ is complete. $A\triangle B$ is defined as $(A\smallsetminus B)\cup(B\smallsetminus A)$. I've tried to prove this. So far, I've proved $\varnothing,X\in\overline{\mathcal{E}}$, or in fact that $\mathcal{E}\subseteq\overline{\mathcal{E}}$, since for any $A\in\mathcal{E}$ we have $A\triangle A=\varnothing$ and obviously $\mu(\varnothing)=0$, and that, supposing $\overline{\mu}$ is well-defined, it extends $\mu$, as one possible $B$ set for which $\overline{\mu}(A)=\mu(B)$ is $A$ itself, as used in the short proof above. My doubts are:

1) Give $A,B\in\overline{\mathcal{E}}$, are $A\cup B$ and $A\smallsetminus B$ elements of $\overline{\mathcal{E}}$? In other words, is $\overline{\mathcal{E}}$ a $\sigma$-algebra?

2) Is it true that for any two such sets $B,B'$ as enter the definition of $\overline{\mu}$, we have that $\mu(B)=\mu(B')$? In other words, is $\overline{\mu}$ well-defined?

3) Is this actually a measure, i.e. a positive $\sigma$-additive function?

4) Is the resulting space $(X,\overline{\mathcal{E}},\overline{\mu})$ actually complete?

  • $\begingroup$ Yes, to all questions. $\endgroup$ – Michael Greinecker Mar 15 '14 at 11:32
  • $\begingroup$ The moment I see this as a definition of the completion of a measure space, I have to assume this is the case. However, an assumption is not satisfactory. That's why I came here: my question is asking for a proof, not just for an answer. $\endgroup$ – MickG Mar 15 '14 at 11:39
  • $\begingroup$ ... and you should show your own attempts at these proofs. $\endgroup$ – GEdgar Mar 15 '14 at 13:33
  • $\begingroup$ @GEdgar in fact I have. I haven't managed much, and what I've managed is in the questions. Any attempt (if indeed there were any starting ideas, which I can't recall) at proving the points in the list was a failure and I deemed it useless to type them into the question because I always found a reason to chuck them away as inconclusive, not just abandon them due to my inability to go anywhere. $\endgroup$ – MickG Apr 23 '14 at 16:28

We first show that the elements of $\bar{\mathcal{E}}$ are exactly the elements that can be written as $A\cup N$ with $A\in\mathcal{E}$ and $N$ being a subset of a measurable set with measure $0$.

For one direction, observe that $A\triangle A\cup N=(A-A\cup N)\cup(A\cup N-A)=(A\cup N-A)\subseteq N$ and $N$ is by assumption a subset of a set with measure $0$. Now assume that $A\triangle B\subseteq C$, $B$ is measurable and $C$ has measure zero. Then $(A-B)\cup (B-A)\subseteq C$. Now $A-B\subseteq C$ implies $A\subseteq B\cup C$ and $B-A\subseteq C$ implies $B-C\subseteq A$. Hence $A=(B-C)\cup (C\cap A)$. Clearly, $B-C$ is measurable and $A\cap C$ is a subset of the measure zero set $C$.

Now let $(A_n)$ be a sequence of sets in $\mathcal{E}$ and $(N_n)$ a sequence of sets such that for some sequence of measurable sets $(C_n)$ we have $N_n\subseteq C_n$ and $\mu(C_n)=0$. Then $\bigcup_n (A_n\cup N_n)=\bigcup_n A_n\cup\bigcup_n N_n$. Since $\bigcup_N N_n\subseteq\bigcup_{n} C_n$ and $\mu(\bigcup_n C_n)\leq\sum_n \mu(C_n)=0$, we see that $\bar{\mathcal{E}}$ is closed under countable unions. Moreover, let $A$ be measurable and $N\subseteq C$ with $C$ being a measurable set of measure zero. Then $X-(A\cup N)= (X-(A\cup C))\cup (C\cap (X-(A\cup N)))$. Since $X-(A\cup C)$ is measurable and $C\cap (X-(A\cup N))$ is a subset of the measurable subset of measure zero $C$, we see that $\bar{\mathcal{E}}$ is closed under complements. Also, we obviously have $\emptyset,X\in\bar{\mathcal{E}}$, so $\bar{\mathcal{E}}$ is a $\sigma$-algebra.

We now show that $\bar{\mu}$ is well defined. So take some $A\subseteq X$ and $B_1,B_2,C_1,C_2$ be measurable sets such that $C_1,C_2$ have measure zero and $A\triangle B_1\subseteq C_1$ and $A\triangle B_2\subseteq C_2$. We have $(A-B_1)\cup (B_1-A)\subseteq C_1$ and $(A-B_2)\cup (B_2-A)\subseteq C_2$. Now $B_1-A\subseteq C_1$ implies $B_1-C_1\subseteq A$ and $A-B_2\subseteq C_2$ implies $A\subseteq B_2\cup C_2$. So we have $B_1-C_1\subseteq B_2\cup C_2$. Since $C_1$ and $C_2$ have measure zero, we get $\mu(B_1)\leq\mu(B_2)$. Similarly, we can show that $\mu(B_2)\leq\mu(B_1)$ and hence $\mu(B_1)=\bar{\mu}(A)=\mu(B_2)$.

Now if we write an element of $\bar{\mathcal{E}}$ as $A\cup N$ with $A$ measurable and $N$ a subset of a measurable null set, we have $\bar{\mu}(A\cup N)=\mu(A)$ since $A\triangle (A\cup N)$ is a subset of a measure zero set and $A\triangle A=\emptyset$. So countable additivity of $\bar{\mu}$ and its completeness follows in a straightforward manner.

  • $\begingroup$ OK my problem is solved. Now since this may be read by quite a few classmates of mine, I'd like to add 2 or 3 details. 1) The first given equality is due to $A\smallsetminus(A\cup N)$ being $\varnothing$ since $A\subseteq A\cup N$; 2) $A\subseteq B\cup C$ and $B\smallsetminus C\subseteq A$ mean every point of $A$ is either one of $B$ or one of $C$ (or both); if you take $B\smallsetminus C$, and take it away from $A$, what you are left with are necessarily all points of $C$, implying $A=(B\smallsetminus C)\cup(C\cap A)$; $\endgroup$ – MickG Mar 15 '14 at 14:03
  • $\begingroup$ 3) The "strangest-looking" (least intuitive) equality is $$X-(A\cup N)=(X-(A\cup C))\cup(C\cap(X-(A\cup N)));$$ let's say $A^c=X\smallsetminus A$ and rewrite that as $$(A\cup N)^c=(A\cup C)^c\cup(C\cap(A\cup N)^c));$$ the second set on the right is clearly a subset of the complement on the left, and the first one is too, since $(A\cup C)^c=A^c\cap C^c\subseteq A^c\cap N^c=(A\cup N)^c$ being $N\subseteq C$; we can further rewrite the equality as $$(A\cup N)^c\smallsetminus C=(A\cup C)^c;$$ $\endgroup$ – MickG Mar 15 '14 at 14:10
  • $\begingroup$ a definition of $A\smallsetminus B$ is $A\cap B^c$; so we can further rewrite this as $$(A^c\cap N^c)\cap C^c=A^c\cap C^c;$$ this clearly holds, since by associativity of the $\cap$ operator the left-hand side can be rewritted as $$(A^c\cap N^c)\cap C^c=A^c\cap(N^c\cap C^c)=A^c\cap C^c,$$ since $C^c\subseteq C^c$ as said above; 4) Countable additivity is straightforward; let $E_n$ be a sequence of elements of our sigma-algebra; their union is $\bigcup_nE_n=(\bigcup_nA_n)\cup(\bigcup_nN_n)$, where for any $n$ we put $E_n=A_n\cup N_n$, $\endgroup$ – MickG Mar 15 '14 at 14:15
  • $\begingroup$ where $A_n$ is measurable and $N_n$ is a subset of $C_n$ with $\mu(C_n)=0$, just as written above. So: $$\bar{\mu}\left(\bigcup_nE_n\right)=\bar{\mu}\bigg(\big(\bigcup_nA_n\big) \cup\big(\bigcup_nN_n\big)\bigg)=\mu\left(\bigcup_nA_n\right)=\sum_{n=1}^\infty \mu(A_n)=\sum_{n=1}^\infty\bar{\mu}(A_n\cup N_n)=\sum_{n=1}^\infty\bar{\mu}(E_n),$$ using the fact the union of $N_n$ is a subset of the union of $C_n$ which has measure zero (again, see above) and the countable additivity of $\mu$ and what you have proved for $\bar{\mu}(A\cup N)$ with $A,N$ as the $A_n,N_n$ above. $\endgroup$ – MickG Mar 15 '14 at 14:17

The questions are easily answered if one has some intuition for small symmetric differences (or other sets containing them) taking the role of exceptions. So I will take this question as an opportunity to explain that intuition and then I will demonstrate how it applies to the questions.

SECTION: The Intuition for Exceptions and Exception Management

If the following three hold:

A1. Most dogs in the kingdom love barking

A2. Most dogs in the kingdom can read.

A3. Each dog in the kingdom who can read and loves barking loves cats.

then the following holds:

A4. Most dogs in the kingdom love cats

That does not seem to require any explanation, but if we are to explain that using sets, we can say that A1 is simply saying that there is a very small subset $E_1$ of the universal set $X$ (= the set of all the dogs in the kingdom) such that all those untypical dogs (in the kingdom) who do NOT love barking are in in $E_1$. And likewise, A2 is simply saying that all the dogs in the kingdom who CANNOT read belong to a small subset $E_2$. Notice how $E_1$ and $E_2$ represent exceptions.

All dogs in the kingdom love barking, except for those dogs in $E_1$. All dogs the kingdom can read, except for those dogs in $E_2$. Hence, all dogs in the kingdom except for those dogs in $E_1 \cup E_2$ have the property of loving barking and being able to read (and hence also the property of loving cats). If you judge $E_1 \cup E_2$ to be a small set, then you would say that most dogs in the kingdom love cats. Is it reasonable to judge the union to be a small set? Yes because it is a union of two small sets.

There is a crucial observation to be made about this argument. It is that the argument consists of two parts: 1. managing exceptions (such as judging that $E_1 \cup E_2$ is a small set), and 2. deduction on typical dogs unburdened by having to think about any exceptional dogs. We were reasoning about typical dogs, i.e., the dogs NOT in $E_1 \cup E_2$, and deducing that those dogs love cats, and when we were doing that, we did NOT worry about exceptional dogs.

Proofs of yes answers to your questions will usually take that form as well: consisting of exception management and deduction on non-exceptional things.

SECTION: Symmetric Difference and Exceptions

Let's assume the kingdom has a bad king who happens to be paranoid. The king announces that all dogs suspected of unkingdom activities should be arrested. The king claims there is an easy test to figure out who is unkingdom and who is not. The test simply checks whether the subject can read and write. Any dog who can read and write is highly likely to be an intellectual and therefore an unkingdom dog, says the king. There are of course false positives and false negatives for this test but those exceptions are rare, says the king.

The king's claim can be expressed in terms of symmetric difference. If $A$ is the set of all dogs in the kingdom who can read, and $B$ the set of all dogs in the kingdom who can write, and $C$ the set of all unkingdom dogs in the kingdom, then if the test were a perfect test, that is, if the test has NO false positive and false negative, then $A \cap B = C$ would hold, or equivalently, $(A \cap B) \Delta C = \emptyset$.

But the king admits that the test may not be perfect. He claims only that the exceptions (false positive and false negative) are rare. So what he's saying is that $A \cap B$ is a good approximation for the target set $C$, in other words, $(A \cap B) \Delta C$ is a small subset of $X$.

This analysis already shows a pattern of how the notion of symmetric difference is usually used in probability theory and measure theory. The pattern is that there is a target set (or a family of target sets) you want to reason about, but instead of directly dealing with the target set, you come up with an easier set that approximates the target set, and then you see how small the symmetric difference between the easier set and the target set is.

Some Convention

Before we go to the questions, let's establish a useful convention this post follows: symbols based on the capital letter E, such as $E, E_1, E_2, E', \ldots$, will always denote sets that are intended to represent exceptions, because we want E to stand for exception.

SECTION: How to Discover a Proof of Yes to the First Question

Suppose that $A, B \in \overline{\mathcal E}$, how do we show that $A \cup B$ is also in $\overline{\mathcal E}$?

At least we know there is $A' \in \mathcal E$ approximating $A$ in the sense that $A' \Delta A$ is so small that it is contained in some null $E_A \in \mathcal E$. Also, there is $B' \in \mathcal E$ approximating $B$ in the sense that $B' \Delta B$ is so small that it is contained in some null $E_B \in \mathcal E$.

We want to reason about the target set $A \cup B$, but is $A' \cup B'$ a good approximation to it? Yes, and that is because $(A \cup B) \Delta (A' \cup B')$ is so small that it is contained in $E_A \cup E_B$ which one can easily check is a null set in $\mathcal E$.

If you don't see why $(A \cup B) \Delta (A' \cup B') \subset E_A \cup E_B$ holds, rather than struggling with a Venn diagram, maybe think this way:

A1. $1_A(x) = 1_{A'}(x)$ holds for all $x$ except for those in $E_A$. (similar to: most dogs in the kingdom love barking)

A2. $1_B(x) = 1_{B'}(x)$ holds for all $x$ except for those in $E_B$ (similar to: most dogs in the kingdom can read)

A3. For each $x \in X$ satisfying $1_A(x) = 1_{A'}(x)$ and $1_B(x) = 1_{B'}(x)$, we have $1_{A' \cup B'}(x) = 1_{A \cup B}(x)$ and $1_{A' \cap B'}(x) = 1_{A \cap B}(x)$ and $1_{A' \setminus B'}(x) = 1_{A \setminus B}(x)$.. (Each dog in the kingdom who can read and love barking loves cats.)


A4. $1_{A' \cup B'}(x) = 1_{A \cup B}(x)$ and $1_{A' \cap B'}(x) = 1_{A \cap B}(x)$ and $1_{A' \setminus B'}(x) = 1_{A \setminus B}(x)$ hold for all $x$ except for those in $E_A \cup E_B$. (similar to: therefore, most dogs in the kingdom love cats)

That means, $(A \cup B) \Delta (A' \cup B') \subset E_A \cup E_B$ holds and $(A \cap B) \Delta (A' \cap B') \subset E_A \cup E_B$ and $(A \setminus B) \Delta (A' \setminus B') \subset E_A \cup E_B$.

What do we know about our approximations $A' \cup B', A' \cap B', A' \setminus B'$? We know they are all in $\mathcal E$ because $\mathcal E$ is a sigma algebra. Since we have shown that they are good enough approximations to our target sets $A \cup B, A \cap B, A \setminus B$, these target sets are all in $\overline{\mathcal E}$.

It helps to think of these sets as events as in probability theory rather than circles in Venn diagrams: A1. the event $A$ occurred if and only if the event $A'$ occurred, unless the exception event $E_A$ occurred. and so on and A4. therefore, the event $A \setminus B$ is equivalent to event $A' \setminus B'$, unless the exception event $E_A \cup E_B$ occurred. Trouble with Venn diagrams when it comes to symmetric difference is that it is hard to draw a Venn diagram involving more than three sets, and even if you manage to draw such a diagram, it does not encourage you to isolate exception management and exception-free deduction.

SECTION: How to Make the Proof Less Verbose

Maybe you can explicitly extract out a lemma like $(A \setminus B) \Delta (A' \setminus B') \subset (A \Delta A') \cup (B \Delta B')$, etc and then rewrite the proof to use such lemmas. Extracting a lemma can be good also because you can reuse the lemma later.

SECTION: How to Discover Proof of Yes to the Second Question

Suppose $A$ is approximated by $A' \in \mathcal E$ so that $A \Delta A'$ is a subset of a null $E' \in \mathcal E$. Suppose it is also approximated by a different $A'' \in \mathcal E$ so that $A \Delta A''$ is a subset of a null $E'' \in \mathcal E$. Now how should we show that $\mu(A')$ = $\mu(A'')$?

Without filling in the details of the proof yet, one can see that the rough sketch of its proof should look like the following: Since $A$ is close to both $A'$ and $A''$, the two sets $A'$ and $A''$ must be close to each other, in fact, they are so close that $\mu(A')$ end up being the same as $\mu(A'')$.

The bad king claims that the literate dogs and the dogs who are intellectuals are same, except for rare exceptional cases, and that the dogs who are intellectuals and the unkingdom dogs are same except a few cases, and therefore the literate dogs and the unkingdom dogs are largely the same. Here the set of dogs who are intellectuals is like the set $A$

$A$ is close to $A'$ in the sense that $1_A = 1_{A'}$ holds for all $x$ except for those in $E'$, and $A$ is close to $A''$ in the sense that $1_A = 1_{A''}$ holds for all $x$ except for those in $E''$, Therefore, $1_{A'} = 1_A = 1_{A''}$ holds for all except for those in $E := E' \cup E''$ which one can check is a null set in $\mathcal E$. So $A'$ and $A''$ are so close to each other that $A' \Delta A''$ is contained in the null $E \in \mathcal E$. Since $A' \Delta A''$ is an element of $\mathcal E$, it is legal to consider $\mu(A' \Delta A'')$ which is of course zero. Then the conclusion $\mu(A')$ = $\mu(A'')$ is just one Venn diagram away.

SECTION: Again, Extracting Lemmas

$(A' \Delta A'') \subset (A' \Delta A) \cup (A \Delta A'')$ (similar to the inequality $|a' - a''| \le |a' - a| + |a - a''|$ in its form and its use).

Another useful lemma you can extract is that $\mu(A')$ = $\mu(A'')$ holds if $A' \Delta A''$ is contained a null set. You will reuse this lemma in a solution to the third question.

SECTION: Third Question

It comes down to proving that if the sets $A_i \in \mathcal E$, ($i \in \mathbb N$) are a.e. disjoint (in the sense that $A_i \cap A_j$ is a null set for each $i,j$ with $i \neq j$) then show that $\sum_i \mu(A_i) = \mu (\bigcup_i A_i)$. We'd wish the sets $A_i$ were disjoint, because then the job would trivial, but that is not true in general, unfortunately. Fortunately, that wish suggests a direction, which is that we should approximate each $A_i$ by some $A_i' \in \mathcal E$ in such a way that the sets $A_i'$ are disjoint and also in a way that the approximations are good enough to conclude $\mu(A_i) = \mu(A_i')$ and $\mu (\bigcup_i A_i) = \mu (\bigcup_i A_i')$.

What points should we add to $A_i$ and what points should we remove from it so that we obtain $A_i'$ with the desired disjointness property? Adding points is not going to help acquire disjointness. So we are going to only remove some points from $A_i$ to define $A_i'$. Now the question is what points should we remove? What were the offending points that prevented us from assuming that the sets $A_i$ were disjoint? The offending points were all the points in the sets $A_i \cap A_j$.That's it. We define $A_k' = A_k \setminus E$ where $E$ is the union of the $n(n-1)$ sets $A_i \cap A_j$.


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