Measure of Limit of Increasing Sequence of Measurable Sets

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $E \in \Sigma$.

Let $\sequence {E_n}_{n \mathop \in \N}$ be an increasing sequence of $\Sigma$-measurable sets such that:


 * $E_n \uparrow E$

where $E_n \uparrow E$ denotes limit of increasing sequence of sets.

Then:


 * $\ds \map \mu E = \lim_{n \mathop \to \infty} \map \mu {E_n}$

Proof
We define a sequence $\sequence {F_n}_{n \mathop \in \N}$ inductively.

Set $F_1 = E_1$.

For $n > 1$, define:


 * $F_n = E_n \setminus E_{n - 1}$

From the definition of set difference we have:


 * $E_n \setminus E_{n - 1} = E_n \cap \paren {X \setminus E_{n - 1} }$

Since $\Sigma$ is closed under complementation, we have:


 * $X \setminus E_{n - 1} \in \Sigma$

and so, from Sigma-Algebra Closed under Countable Intersection:


 * $F_n \in \Sigma$

for $n > 1$.

Clearly $F_1 \in \Sigma$ also.

Since $\sequence {E_n}_{n \in \N}$ is increasing, we have that:


 * $E_i \subseteq E_j$ for $i \le j$.

So:


 * $F_n$ is disjoint to $E_1, E_2, \ldots, E_{n - 1}$.

Since $F_i \subseteq E_i$ for each $i$, we have:


 * $F_n$ is disjoint to $F_1, F_2, \ldots, F_{n - 1}$.

Picking $n = \max \set {i, j}$ we obtain that:


 * $F_i \cap F_j = \O$ for $i \ne j$.

We now show that:


 * $\ds E_k = \bigcup_{i \mathop = 1}^k F_i$

We have:


 * $\ds x \in \bigcup_{i \mathop = 1}^k F_i$




 * $x \in F_i$ for some $1 \le i \le k$.

This is equivalent to:


 * $x \in E_i$ for some $1 \le i \le k$.

Since $\sequence {E_n}_{n \in \N}$ increasing sequence, this is equivalent to:


 * $x \in E_k$

showing that:


 * $\ds x \in \bigcup_{i \mathop = 1}^k F_i$ $x \in E_k$

giving:


 * $\ds E_k = \bigcup_{i \mathop = 1}^k F_i$

So, we have:


 * $\ds \bigcup_{k \mathop = 1}^\infty E_k = \bigcup_{i \mathop = 1}^\infty F_i$

That is:


 * $\ds E = \bigcup_{i \mathop = 1}^\infty F_i$

from the definition of a limit of an increasing sequence of sets.

We can now compute: