Consider measure space $(S, \Sigma, \mu) = (\mathbb R, \mathscr B(\mathbb R), \lambda)$.
Let $V^C \subseteq S$ denote the set of all numbers in $[0,1]$ whose decimal representations don't contain the number 7.
Prove that $V^C \in \Sigma$.
Compute $\lambda(V^C)$.
What I tried:
- I think we have
$$V = \bigcup_{n=1}^{\infty} V_n$$
where $v_n \in V_n$ can be written $v_n = 0.s_1s_2...$ where $s_1 \ne 7, ..., s_{n-1} \ne 7, s_n = 7$
$$V_n = \bigcup_{s_1, ..., s_{n-1} \ne 7} [0.s_1...s_{n-1}7, 0.s_1...s_{n-1}8)$$
$V$ is a finite union of pairwise disjoint $9^{n-1}$ intervals and hence is a Borel set, which is Lebesgue measurable.
$V$ is a countable union of pairwise disjoint Borel sets and hence is a Borel set, which is Lebesgue measurable.
Thus, $V^C$ is Lebesgue measurable.
2.
$$\lambda(V) = \lambda(\bigcup_{n=1}^{\infty} V_n) = \sum_{n=1}^{\infty} \lambda(V_n)$$
$$\lambda(V_n) = \lambda \left(\bigcup_{s_1, ..., s_{n-1} \ne 7} [0.s_1...s_{n-1}7, 0.s_1...s_{n-1}8)\right)$$
$$= \sum_{i=1}^{9^{n-1}} \lambda ([0.s_1...s_{n-1}7, 0.s_1...s_{n-1}8))$$
$$= \sum_{i=1}^{9^{n-1}} \frac{1}{10^n} = \frac{1}{10^n} \sum_{i=1}^{9^{n-1}} (1) = \frac{1}{10^n} (9^{n-1} - 1 + 1) = \frac{9^{n-1}}{10^n}$$
$$\to \lambda(V) = \sum_{n=1}^{\infty} \lambda(V_n) = \sum_{n=1}^{\infty} \frac{9^{n-1}}{10^n} = 1$$
$$\therefore, \lambda(V^C) = 0$$
This line $$= \sum_{i=1}^{9^{n-1}} \lambda ([0.s_1...s_{n-1}7, 0.s_1...s_{n-1}8))$$ doesn't immeadiately make sense. Your summands don't depend on i in any way. Use the subscript you had before, ie $$= \sum_{s_1,\ldots, s_{n-1} \ne 7} \lambda ([0.s_1...s_{n-1}7, 0.s_1...s_{n-1}8)).$$ I should say that what you have is correct, but it doesn't look like it follows, and the reader will have to think to realise why it follows. You don't want that, you want the progression of ideas to be obvious.
I'd also be sure to add some explanatory sentences, just so that what you are doing is clear. Less logical symbols, more words.