Here is Prob. 25, Chap. 2, in the book Real Analysis by H. L. Royden and P. M. Fitzpatrick, 4th edition:
Show that the assumption that $m\left( B_1 \right) < \infty$ is necessary in part (ii) of the theorem regarding continuity of measure.
Here is the theorem regarding continuity of measure (i.e. Theorem 15) in Royden:
Lebesgue measure possesses the following continuity properties:
(i) If $\left\{ A_k \right\}_{k=1}^\infty$ is an ascending collection of measurable sets, then $$ m \left( \bigcup_{k=1}^\infty A_k \right) = \lim_{k \to \infty} m \left( A_k \right). $$
(ii) If $\left\{ B_k \right\}_{k=1}^\infty$ is a descending collection of measurable sets and $m \left( B_1 \right) < \infty$, then $$ m \left( \bigcap_{k=1}^\infty B_k \right) = \lim_{k \to \infty} m \left( B_k \right). $$
My Attempt:
For each positive integer $k$, let us put $$ B_k := \big( k, \infty \big) = \big\{ x \in \mathbb{R} \mid k < x < \infty \big\}. $$
Then, for each $k$, we have $$ B_k = \big( k, \infty \big) \supset \big( k+1, \infty \big) = B_{k+1}. $$
And, by Proposition 8 every set $B_k$, being an interval, is measurable, and by Proposition 1 we have $$ m \left( B_k \right) = m^* \left( B_k \right) = l \left( B_k \right) = \infty. $$ Therefore $$ \lim_{k \to \infty} m \left( B_k \right) = \infty. \tag{1} $$
On the other hand, we note that $$ \bigcap_{k=1}^\infty B_k = \emptyset, $$ because for every real number $a$ we can find a positive integer $k > a$, and thus $a \not\in B_k$ for any such $k$; therefore we obtain $$ m \left( \bigcap_{k=1}^\infty B_k \right) = 0. \tag{2} $$
From (1) and (2) we conclude that part (ii) of Theorem 15 does not hold for this particular countable collection of measurable sets for which $m \left( B_1 \right) \not< \infty$.
Is my proof correct and clear enough? If not, then where are the issues?