Proving if $E_n $ is an increasing sequence in $\sum$ then $\mu(\cup^\infty_{n=1} E_n) =$ lim $\mu$ $ (E_n)$ [duplicate]

Question

Let $\mu$ be a measure defined on a $\sigma$ algebra $\sum$. I am trying to prove if $E_n $ is an increasing sequence in $\sum$ then $\mu(\cup^\infty_{n=1} E_n) = $ lim $\mu $ $(E_n)$

This is the proof.

We assume that $\mu(E_n) < +\infty$.

Let $A_1 = E_1$ and $A_n = E_n \backslash E_{n-1}$ for $n \geq 1$

Then $A_n$ is a disjoint sequence of sets in $\sum$ such that $E_n = \cup^n _{j=1} A_j$ and $\cup^\infty_{E_n} = \cup^\infty_{n=1} A_n$

Why do we have to introduce another sequence $A_n$?

$\mu$ is countable additive so $\mu(\cup^\infty_{n=1} E_n)=\sum^\infty_{n=1} \mu(A_n)$.

Why does this last step equal lim$\sum^m_{n=1} \mu (A_n)$?

How does the proof then jumps to saying $\mu(A_n) = \mu(E_n) - \mu(E_{n-1})$ and hence $\sum^m_{n=1} \mu(A_n) = \mu (E_m)$?

Answer 1

Why do we have to introduce another sequence $A_n$?

The $E_n$ are not disjoint, and countable additivity does not hold unless you have a disjoint collection of sets.

$\mu$ is countable additive so $\mu\left(\cup_{n=1}^\infty E_n\right) = \sum_{n=1}^\infty \mu(A_n)$.

Why does this last step equal $\lim \sum_{n=1}^m \mu(A_n)$?

The step missing here is to notice that $\bigcup_{n=1}^\infty E_n = \bigcup_{n=1}^\infty A_n$. So $\mu \left( \bigcup_{n=1}^\infty E_n \right)=\mu \left(\bigcup_{n=1}^\infty A_n \right)$, and since the $A_n$ are disjoint, countable additivity gives $\mu \left(\bigcup_{n=1}^\infty A_n \right)=\sum_{n=1}^\infty \mu(A_n)$. Finally, $\sum_{n=1}^\infty \mu(A_n) = \lim_{m \to \infty} \sum_{n=1}^m \mu(A_n)$ by the definition of an infinite series.

How does the proof then jumps to saying $\mu(A_n)=\mu(E_n)−\mu(E_{n−1})$ and hence $\sum_{n=1}^m \mu(A_n) = \mu (E_m)$?

Notice that $E_n = A_n \cup E_{n-1}$, and $A_n,E_{n-1}$ are disjoint (for $n \geq 2$). So the additivity of $\mu$ gives $\mu(E_n) = \mu(A_n) + \mu(E_{n-1})$, and thus $\mu(A_n) =\mu (E_n)- \mu(E_{n-1})$.

This gives you a telescoping sum: $$\sum_{n=1}^m \mu(A_n) = \mu(A_1) + \sum_{n=2}^m \mu(A_n) = \mu(E_1) + \sum_{n=2}^m(\mu(E_n)-\mu(E_{n-1}))= \mu(E_m).$$