A counter-example for how the Downward Monotone Convergence theorem can fail is given by; Let $n = 1$ so we are working in $\mathbb{R}^1$ and let $E_n = [n,\infty)$. Then $m(E_n) = \infty$ for all $n$. And $E_n$ satisfies the criterion $E_1 \supset E_2 \supset E_3 \supset \cdots$. However, $m(\cap_{n=1}^{\infty} E_n) = m(\emptyset) = 0 \neq \infty = \lim \, m(E_n)$.
So let's go over the proof of the theorem to see exactly where the finiteness of an $E_n$ is used. First, using the same idea from the first problem we can write each $E_n$ as a disjoint union:
$$
E_n = \cup_{k=n}^{\infty} F_n \, \bigcup \, S
$$
Where each $F_n$ is given by $F_n = E_n - E_{n+1}$, and $S$ is given by $S = \cap^{\infty} E_n$. It should be clear that these sets are mutually disjoint. Now the additivity of measures gives
$$
m(E_n) = m(S) + \Sigma_{k=n}^{\infty} m(F_k)
$$
The statement that $m(E_n)$ is finite for some $n$ is exactly the statement that $\Sigma_{k=n}^{\infty} m(F_k)$ converges for some $n$ and that $m(S)$ is finite.
That $m(S)$ is finite is immediate from the monotonicity of measures and that $S \subset E_n$ for all $n$ along with that $E_n$ has finite measure for some $n$. And since the series $\Sigma_{k=n}^{\infty} m(F_k)$ converges for some $n$ we have:
$$
\lim_n \, \Sigma_{k=n}^{\infty} m(F_k) = 0.
$$
We get:
$$
\lim_n \, m(E_n) = m(S) + \lim_n \, \Sigma_{k=n}^{\infty} m(F_k) = m(S).
$$
Yes. Suppose that ${\frak m}^\ast E = 0$. On one hand, the inequality: $${\frak m}^\ast T \leq {\frak m}^\ast(T \cap E) + {\frak m}^\ast(T \cap E^c)$$always holds. On the other hand: $${\frak m}^\ast(T \cap E) \leq {\frak m}^\ast E = 0 \implies {\frak m}^\ast(T \cap E) = 0, \quad {\frak m}^\ast(T \cap E^c) \leq {\frak m}^\ast T$$gives the other inequality.
Best Answer
The theorem is true for measurable sets.
Proof for the general case:
There is a subsequence { $ E_k $} such that $\mu^* (E_{k+1} ) - \mu^* (E_k ) \le \frac {\epsilon}{2^{k+1}} $
Lets first construct such subsequence :
$ \lim\limits_{n\mapsto \infty}\mu^*(E_n) \ge \mu^*(E_{n+1}) \ge \mu^*(E_n)$
Choose $E_1$ such that $ \lim\limits_{n\mapsto \infty}\mu^*(E_n) -\mu^*(E_1) \le \frac {\epsilon}{2} $
Choose $E_2$ such that $E_1 \subseteq E_2$ and $ \lim\limits_{n\mapsto \infty}\mu^*(E_n) -\mu^*(E_2) \le \frac {\epsilon}{2^3} $
Choose $E_3$ such that $E_2 \subseteq E_3$ and $ \lim\limits_{n\mapsto \infty}\mu^*(E_n) -\mu^*(E_3) \le \frac {\epsilon}{2^4} $ Then use induction
Step 1: cover $E_k$ with union of open intervals $\bigcup_{i=1}^\infty I_i = L_k $ Such that $ \mu^* (L_k) \le \mu^* (E_k) + \frac {\epsilon}{2^k} $
By Caratheodory condition $ \mu^* (E_{k+1} \bigcap L_k^c ) = \mu^* (E_{k+1} ) - \mu^* (E_{k+1} \bigcap L_k ) $ $\mu^*(E_k) \le \mu^*(E_{k+1} \bigcap L_k ) \le \mu^*(L_k) $
Therefore $ \mu^*(E_{k+1} \bigcap L_k^c ) \le \mu^*(E_{k+1}) - \mu^* (E_k) \le \frac {\epsilon}{2^{k+1}}$
Step 2 : Let $ G_{k+1} = E_{k+1} \bigcap L_k^c $
$ L_k \bigcup G_{k+1}$ contains $E_{k+1}$
Now cover $ L_k \bigcup G_{k+1}$ with union of intervals $\bigcup_{i=1}^\infty I_i = H_{k+1} $ Such that $\mu^* (H_{k+1}) \le \mu^*( L_k \bigcup G_{k+1}) + \frac {\epsilon}{2^{k+1}} $
$\mu^* (H_{k+1}) \le \mu^*( L_k) + \mu^*( G_{k+1}) \le \mu^*(E_k) + \frac { \epsilon}{2^k} +\frac { 2\epsilon}{2^{k+1} }$
Now using $H_{k+1} $ as the cover for $ E_{k+1} $
As seen $\mu^*(E_{k+1}) \le \mu^*(H_{k+1})$ $\le \mu^*(E_{k+1})$ $+ \frac {\epsilon}{2^k}$ $+\frac {2\epsilon}{2^{k+1}} $
Now apply step 1 and 2 to $ E_{k+1}$ and $ E_{k+2}$ and get :
$ \mu^*(E_{k+2}) \le \mu^* (H_{k+2}) \le \mu^*(E_{k+2}) + \frac { \epsilon}{2^k} +\frac { 2\epsilon}{2^{k+1} } +\frac { 2\epsilon}{2^{k+2} }$
It is obvious $ E \subseteq \bigcup_{i=1}^\infty H_k $
$H_k \subseteq H_{k+1}$
$ \mu^*(E_{k}) \le \mu^* (H_{k}) \le \mu^*(E_{k}) + 4 \epsilon $
Notice that the theorem is valid for $H_k$ as it is a measurable set (union of intervals) So $\lim\limits_{k\mapsto \infty} \mu^*(H_k) \ge \mu^*(E) $
$\lim\limits_{k\mapsto \infty}\mu^*(E_{k}) \le \lim\limits_{k\mapsto \infty} \mu^*(H_k) \le \lim\limits_{k\mapsto \infty}\mu^*(E_{k}) + 4\epsilon $
$\lim\limits_{k\mapsto \infty}\mu^*(E_{k}) \le \mu^*(E) \le \lim\limits_{k\mapsto \infty}\mu^*(E_{k}) + 4\epsilon $
Because $\epsilon $ is arbitrary the proof is complete.
Remark: The proof is straight forward for measurable sets but not so for arbitrary sets. Actually it is true and it was given as an exercise in 'The Integrals of Lebesgue, Denjoy , Perron , and Henstock (Graduate Studies in Mathematics Volume 4 )' by Russell A. Gordon . It is theorem 1.15 in the book.
This theorem allows the short proofs of Dominated convergence theorem, Vitali Convergence Theorem, Monotone Convergence Theorem , Egorov's theorem and Luzin's theorem without dwelling much on the machinery of measure theory.