User:Caliburn/s/mt/Measure of Limit of Decreasing 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 decreasing sequence of $\Sigma$-measurable sets such that:


 * $E_n \downarrow E$

where $E_n \downarrow E$ denotes limit of decreasing sequence of sets, and:


 * there exists $m \in \N$ such that $\map \mu {E_m}$ is finite.

Then:


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

Proof
First suppose that $\map \mu {E_1}$ is finite.

Now for each $k$ define:


 * $C_k = E_1 \setminus E_k$

From the definition of set difference, we have:


 * $E_1 \setminus E_k = E_1 \cap \paren {X \setminus E_k}$

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


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

From Set Complement inverts Subsets, we have:


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

So from Set Intersection Preserves Subsets:


 * $E_1 \cap \paren {X \setminus E_i} \subseteq E_1 \cap \paren {X \setminus E_j}$ for $i \le j$

giving:


 * $C_i \subseteq C_j$ for $i \le j$.

That is:


 * $\sequence {C_k}_{k \in \N}$ is an increasing sequence.

Further, we have:

So from Measure of Limit of Increasing Sequence of Measurable Sets, we have:


 * $\ds \map \mu {E_1 \setminus E} = \lim_{k \mathop \to \infty} \map \mu {C_k}$

That is:


 * $\ds \map \mu {E_1 \setminus E} = \lim_{k \mathop \to \infty} \map \mu {E_1 \setminus E_k}$

Since:


 * $\ds E_1 \subseteq \bigcup_{i \mathop = 1}^\infty E_i = E$

we have:


 * $\ds \map \mu {E_1 \setminus E} = \map \mu {E_1} - \map \mu E$

from Measure of Set Difference with Subset.

Similarly, we have:


 * $E_k \subseteq E_1$

since $\sequence {E_k}_{k \in \N}$ is a decreasing sequence.

So, applying Measure of Set Difference with Subset again:


 * $\ds \map \mu {E_1 \setminus E_k} = \map \mu {E_1} - \map \mu {E_k}$

Since:


 * $\map \mu {E_1} < \infty$

we also obtain that:


 * $\map \mu {E_k} < \infty$

for each $k$, from Measure is Monotone.

Hence:


 * $\ds \map \mu {E_1} - \map \mu E = \lim_{k \mathop \to \infty} \paren {\map \mu {E_1} - \map \mu {E_k} }$

So, from Combination Theorem for Sequences: Real: Sum Rule, we have:


 * $\ds \map \mu E = \lim_{k \mathop \to \infty} \map \mu {E_k}$

Now suppose that any $\map \mu {E_m}$ is finite.

Then $\sequence {F_k}_{k \in \N}$ with:


 * $F_k = E_{m + k - 1}$

for each $k$ is a decreasing sequence of $\Sigma$-measurable sets with:


 * $\map \mu {F_1} = \map \mu {E_m} < \infty$

We also have:


 * $\ds \bigcap_{k \mathop = 1}^\infty F_k = \bigcap_{k \mathop = 1}^\infty E_{m + k - 1} = E$

so:


 * $F_k \downarrow E$

We therefore obtain:


 * $\ds \map \mu E = \lim_{k \mathop \to \infty} \map \mu {F_k}$

That is:


 * $\ds \map \mu E = \lim_{k \mathop \to \infty} \map \mu {E_k}$