Suppose $E_1= [1, 1 \frac12] , E_2 = (2, 2\frac14), E_3 = [3, 3\frac18], E_4 = (4 , 4 \frac{1}{16}) , \dots , E= \bigcup_{n=1}^{\infty}E_n $
i) Show $E$ is measurable
ii) Compute $m(E)$
Here is what I know from reading: If $E$ is an element in $M$, we say $E$ is Lebesgue measurable and $M$ will be called Lebesgue measure.
I was looking at this problem in a book which explained some things about Lebesgue measures in my self study class. I asked my TA but he said he would not do this since we are just learning some proofs. This topic was also not covered.
I would like to see how we would solve a numerical proof like this. I did not do anything like this yet. It seems like a nice proof of a problem to know for other problems.
$E$ is the countable union of intervals, which are measurable . Therefore, since the Lebesgue-measurable sets are a $\sigma$-algebra, we conclude that countable unions of measurable sets are measurable, i.e. $E$ is measurable.
Now, regarding the second question, all sets $E_i$ are disjoint, therefore from countable additivity of the Lebesgue measure: $$m(E)=m\left(\bigcup\limits_{n=1}^\infty E_n\right)=\sum\limits_{n=1}^\infty m(E_n)=$$ $$=m(E_1)+m(E_2)+\dots=$$ $$=\frac12+\frac14+\frac18+\dots=$$ $$=1$$
where $m(E_1)=1\frac12-1=\frac12$, $m(E_2)=2\frac14-2=\frac14,\dots$