Suppose $\{E_n\}$ are measurable and suppose we have the following lemma:
If in addition, $E_n \subseteq E_{n+1}$,
then lim$_{n \rightarrow \infty}$ $m(E_n)$ = $m(\bigcup_{n=1}^{\infty} E_n)$.
Then, prove:
If $E_n$ measurable, $m(E_1) < + \infty$ AND $E_{n+1} \subseteq E_n$,
THEN, lim$_{n \rightarrow \infty}$ $m(E_n)$ = $m(\bigcap_{n=1}^{\infty}E_n)$
So, Since the $E_n$ are measurable, we have monotonicity,
i.e.,
$E_{n+1} \subseteq E_n$ $\Rightarrow$ $m(E_{n+1}) \leq m(E_n)$
but $m(E_1) < + \infty$ gives us
$m(E_{n+1}) \leq m(E_n) = k$ ; $\forall n \in \mathbb{Z}^+$ and SOME $k \in \mathbb{R}$ (this denotes measure of $E_1$).
I get stuck on how to apply the Lemma, am I supposed to use the analogous version of the Lemma?
i.e., replacing $n$ with $n+1$?
and would this give
lim$_{n \rightarrow \infty}$ $m(E_{n+1})$ = $m(\bigcup_{n=1}^{\infty} E_{n+1})$??
or do I rewrite
$m(E_{n+1}$) = $m(E_n$) + $m(E_1)$
I highly doubt this, I never make assumptions in proofs I am not positive about and I never say I get something till it fully makes sense to me, please help, you can see where I am stuck. I apologize in advance if I am missing the obvious. I am merely a noob at measure theory.
Best Answer
$E_1\setminus E_n$ increases to $E_1\setminus \cap_n E_n$, so $m(E_1\setminus E_n) \to m(E_1\setminus \cap_n E_n)$. This gives $m(E_1)-m(E_n) \to m(E_1)-m(\cap_n E_n)$. Cancel $m(E_1)$ to get the result.