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.

Suppose also that $\map \mu {E_1} < \infty$.

Then:


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

Proof
From Relative Complement of Decreasing Sequence of Sets is Increasing, we have:


 * $\sequence {E_1 \setminus E_n}_{n \mathop \in \N}$ is increasing.

Further, we have:

so:


 * $E_1 \setminus E_n \uparrow E_1 \setminus E$

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


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

From Measure of Set Difference with Subset, we have:


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

since $\map \mu {E_1} < \infty$.

From the Difference Rule for Real Sequences, we then have:


 * $\ds \map \mu {E_1} - \map \mu E = \map \mu {E_1} - \lim_{n \mathop \to \infty} \map \mu {E_n}$

giving:


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