Limit of measures is the measure of the intersection

Question

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.

Answer 1

Suppose that $m(E_1) < \infty$, since $E_n \supseteq E_{n+1}$ then by monotoniticty of the Lebesgue measure we have that \begin{align*} m(E_n) < \infty \end{align*} for every $n$. Thus for every $n$ we can define \begin{align*} G_n:=E_1 \setminus E_n \end{align*} Then the $G_n$ are increasing. Note that \begin{eqnarray*} \bigcup_{n=1}^\infty E_1 \backslash E_n &=& \bigcup_{n=1}^\infty (E_1 \cap E_n^c) \\ &=& E_1 \cap (\bigcup_{n=1}^\infty E_n^c) \\ &=&E_1 \cap (\bigcap_{n=1}^\infty E_n)^c \\ &=& E_1 \setminus (\bigcap_{n=1}^\infty E_n) \end{eqnarray*} Furthermore, as the $G_n$ are increasing, we can apply our lemma from above thus \begin{eqnarray*} m(E_1) - m(\bigcap_{n=1}^\infty E_n)&=&m(E_1 \setminus (\bigcap_{n=1}^\infty E_n))\\ &=&m( \bigcup_{n=1}^\infty E_1 \backslash E_n)\\ &=&\lim_{n \rightarrow \infty} m( E_1 \backslash E_n)\\ &=&\lim_{n \rightarrow \infty} [m(E_1) - m(E_n)] \\ &=&m(E_1) - \lim_{n \rightarrow \infty} m(E_n) \end{eqnarray*} But $m(E_1)$ is finite thus we can subtract it from both sides leaving us with $$ m(\bigcap_{n=1}^\infty E_n) = \lim_{n \rightarrow \infty} m(E_n)$$ as needed.

Answer 2

$E_1\setminus E_n$ 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.