CAN ANYONE HELP WITH PART (A)
Let A be the set of numbers in the interval [0, 1] that have the digit “0” in the first, second or third place in their decimal expansion.
(a) Show that A belongs to the σ-algebra generated by the intervals. (Do this by showing that A can be got by applying a countable number of unions/ intersections/complements to suitable intervals.)
(b) Find m(A), where m denotes Lebesgue measure
You first consider the set of numbers with decimal expansion starting with a 0 i.e [0,1) .
Then for each one of the 9 remaining interval [0.1,0.2),[0.2,0.3),…,[0.9,1) [ 0.1 , 0.2 ) , [ 0.2 , 0.3 ) , … , [ 0.9 , 1 ) you then consider the numbers with 0 as a second digit i.e [0.10,0.11),[0.20,0.21),…,[0.90,0.91) [ 0.10 , 0.11 ) , [ 0.20 , 0.21 ) , … , [ 0.90 , 0.91 ) .
And you do it again a third time.
At each step the remaining Lebesgue measure is multiplied by (9/10) therefore after step three (9/10^3)
So, there is a detail that needs to be filled, since, as an example, the number $\frac{1}{10}$ has two decimal expansions, they are $0,1$ and $0,0999...$, then, using the first one, $\frac{1}{10}\notin A$ and the second one, $\frac{1}{10}\in A$. In order to have no contradictions in the construction of $A$, we must fix a cryteria to choose between these two.
This phenomenom of duplicated decimal expansions occurs with numbers of the form $0,a_1a_2a_3...a_n1$, since $$ 0,a_1a_2a_3...a_n1 = 0,a_1a_2a_3...a_n09999999999... $$
What does happen is that any number has an unique infinite decimal expansion (here, by infinite, we mean a decimal expansion that has infinite non-zero algarisms), so, in the construction of $A$, we must consider the infinite decimal expansion of the number.
Let $A_1$ be the set of numbers that have 0 at the first place of their infinite decimal expansion. Then $A_1$ is precisely $\left[0,\frac{1}{10}\right]$, for if $\alpha\in\left[0,\frac{1}{10}\right]$, then the infinite decimal expansion of $\alpha$, say $0,a_1a_2a_3...a_n...$, satisfies $$ 0,a_1a_2a_3...a_n...\leq \frac{1}{10} = 0,09999, $$ and, since $0,a_1a_2a_3...a_n...$ has infinite non-zero algarisms, we must have $a_1=0$.
Now, let $A_2$ be the set of numbers with a non-zero algarism at the first place and $0$ at the second place. First, we get the set of numbers that have $0$ in both first and second place. This set is precisely $\left[0,\frac{1}{100}\right]$ (to conclude that, you just need a similar argument to the above one). Then, the set of numbers with $1$ in the first place and $0$ in the second is $$ \frac{1}{10}+\left[0,\frac{1}{100}\right]. $$ The set of numbers with $2$ in the first place and $0$ in the second is $$ \frac{2}{10}+\left[0,\frac{1}{100}\right], $$ and so on.
Then, $A_2$ is the union of the sets above for each $n\in\{1,...,9\}$, that is $$ A_2=\bigcup_{n=1}^9 \frac{n}{10}+\left[0,\frac{1}{100}\right]. $$
Finally, we must consider the set $A_3$ of numbers that has non-zero algarisms at the first and second places and has $0$ at the third place. The set of numbers that have $0$ at both the first and second places is $\left[0,\frac{1}{1000}\right]$. Then, the set of numbers that have the algarism $a_1$ in the first place, $a_2$ in the second place and $0$ at the third place is $$ 0,a_1a_2+\left[0,\frac{1}{1000}\right]. $$ Then $A_3$ will be the union $$ A_3=\bigcup_{a_1=1}^9\left(\bigcup_{a_2=1}^9 0,a_1a_2+\left[0,\frac{1}{1000}\right]\right). $$
Finally, just note that $A=A_1\cup A_2 \cup A_3$, and each one is the union of closed intervals.
Another way to conclude this, which is faster but harder, is to fix the set $B=\{0,a_1a_2a_3: a_i=0,\mbox{ for some i\in\{1,2,3\}}$ and prove that $$ A=\bigcup_{b\in B} b+\left[0,\frac{1}{1000}\right]. $$