Countable Union of Measurable Sets as Disjoint Union of Measurable Sets

Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

Let $\sequence {A_n}_{n \in \N}$ be a sequence of $\Sigma$-measurable sets.

Then there exists a sequence of $\Sigma$-measurable pairwise disjoint sets $\sequence {B_n}_{n \in \N}$ such that:

$\ds \bigcup_{n \mathop = 1}^\infty A_n = \bigcup_{n \mathop = 1}^\infty B_n$

That is:

any countable union of $\Sigma$-measurable sets can be rewritten as the countable union of pairwise disjoint $\Sigma$-measurable sets.

Proof

Define the sequence $\sequence {B_i}_{i \mathop \in \N}$ by:

$B_1 = A_1$

and:

$\ds B_n = A_n \setminus \paren {\bigcup_{i \mathop = 1}^{n - 1} A_i}$

for $n > 1$.

We show that each $B_n$ is $\Sigma$-measurable.

Since $A_1 \in \Sigma$, we immediately get $B_1 \in \Sigma$.

By De Morgan's Laws, we have:

$\ds B_n = A_n \cap \paren {\bigcap_{i \mathop = 1}^{n - 1} \paren {X \setminus A_i} }$

Note that since $A_i \in \Sigma$ for each $i$, we have:

$X \setminus A_i \in \Sigma$ for each $i$

by the definition of a $\sigma$-algebra.

Then from Sigma-Algebra Closed under Countable Intersection, we have:

$\ds \bigcap_{i \mathop = 1}^{n - 1} \paren {X \setminus A_i} \in \Sigma$

Applying Sigma-Algebra Closed under Countable Intersection again we have:

$\ds B_n = A_n \cap \paren {\bigcap_{i \mathop = 1}^{n - 1} \paren {X \setminus A_i} } \in \Sigma$

for $n > 1$.

So each $B_n$ is $\Sigma$-measurable.

We now show that $\sequence {B_n}_{n \mathop \in \N}$ is a pairwise disjoint family of sets.

Since:

$\ds B_n = A_n \setminus \paren {\bigcup_{i \mathop = 1}^{n - 1} A_i}$

By the definition of set difference, we have:

$\ds B_n \cap \paren {\bigcup_{i \mathop = 1}^{n - 1} A_i} = \O$

so:

$B_n$ is disjoint to $A_1, A_2, \ldots, A_{n - 1}$ for each $n > 1$.

Since $B_i \subseteq A_i$ for each $i$, we therefore have:

$B_n$ is disjoint to $B_1, B_2, \ldots, B_{n - 1}$ for each $n > 1$.

So if $i \ne j$, we have, setting $n = \max \set {i, j}$ and $m = \min \set {i, j}$:

$B_n$ is disjoint to $B_m$.

So:

$\sequence {B_n}_{n \mathop \in \N}$ is a pairwise disjoint family of sets.

That is:

$B_n \cap B_m = B_i \cap B_j = \O$ whenever $i \ne j$.

It remains to show that:

$\ds \bigcup_{n \mathop = 1}^\infty A_n = \bigcup_{n \mathop = 1}^\infty B_n$

Let:

$\ds x \in \bigcup_{n \mathop = 1}^\infty B_n$

then:

$\ds x \in B_n$ for some $n$.

That is:

$\ds x \in A_n \setminus \paren {\bigcup_{i \mathop = 1}^{n - 1} A_i}$ for some $n$.

so:

$x \in A_n$

giving:

$\ds x \in \bigcup_{n \mathop = 1}^\infty A_n$

So:

$\ds \bigcup_{n \mathop = 1}^\infty B_n \subseteq \bigcup_{n \mathop = 1}^\infty A_n$

Now let:

$\ds x \in \bigcup_{n \mathop = 1}^\infty A_n$

then:

$\ds x \in A_n$ for some $n$.

By the well-ordering principle, there exists a least $m$ such that $x \in A_m$.

Then:

$\ds x \not \in \bigcup_{i \mathop = 1}^{m - 1} A_i$

By the definition of set difference, we have:

$\ds x \in A_m \setminus \paren {\bigcup_{i \mathop = 1}^{m - 1} A_i} = B_m$

so:

$\ds x \in \bigcup_{m \mathop = 1}^\infty B_m$

giving:

$\ds \bigcup_{n \mathop = 1}^\infty A_n \subseteq \bigcup_{n \mathop = 1}^\infty B_n$

So we obtain:

$\ds \bigcup_{n \mathop = 1}^\infty A_n = \bigcup_{n \mathop = 1}^\infty B_n$

$\blacksquare$