Lebesgue measure of an intersection of a sequence of subsets This is exercise 1.19 from "A User-Friendly Introduction to Lebesgue Measure and Integration" by Gail S. Nelson, and $m(E)$ is notation for Lebesgue measure of set $E$:
Let ${E_{k}}$ be a sequence of Lebesgue measurable sets with $E_{1}\supseteq E_{2}\supseteq E_{3}\supseteq\ldots$. Define the set $E$ to be $E=\cap_{k=1}^{\infty}E_{k}$. If $m(E_{1})<\infty$, show that $m(E)=\lim_{k\to\infty}m(E_{k})$. Show by example that this need not to be the case if we remove the assumption that $m(E_{1})<\infty$.
My attempt of proof: Since $E$ is subset of $E_{k}$ for each $k$, then $m(E)\leq m(E_{k})$ - this is according to a theorem stating that Lebesgue measure of a subset is less or equal than Lebesgue measure of its superset. Thus, it's certainly true that $m(E)\leq\lim_{k\to\infty}m(E_{k})$ too. If the limit is 0, then $m(E)$ is equal to the limit, as Lebesgue measure of any set is greater or equal to zero.  However, now I don't see how to eventually prove that $m(E)$ cannot be less that the limit, in case when limit greater than 0.  Also, for the second part of the question, I cannot find a counter-example for the case when $m(E_{1})=\infty$.

If  $F_n:=E_n-E_{n+1}$ for $n=1,2,\dots$ then $E_k=E\cup\bigcup_{n=k}^{\infty}F_n$ and the sets in the union are disjoint, so that:
$$m(E_k)=m(E)+\sum_{n=k}^{\infty}m(F_n)$$
So if $m(E_{k_0})<\infty$ for some $k_0$ then $m(E)<\infty$ and $\sum_{n=k_0}^{\infty}m(F_n)<\infty$ and consequently: $$\lim_{k\to\infty}m(E_k)=\lim_{k\to\infty}\left[m(E)+\sum_{n=k}^{\infty}m(F_n)\right]=m(E)+\lim_{k\to\infty}\sum_{n=k}^{\infty}m(F_n)=m(E)$$

Counterexample will only work if $m(E_n)=\infty$ for every $n$. You could take $E_n=[n,\infty)$.
