# Show that $λ(N)=\bigcap_{m=1}^{\infty}(U_m) = 0$

Let $A=\{a_1, a_2, . . .\}$ be a countable subset of R and let λ be the Lebesgue measure on $(\mathbb R,\mathcal M)$. Define for each $k ∈ \mathbb N$

$$U_k =\bigcup_{n=1}^{\infty}(a_n−\frac{1}{2^{n+k}},a_n+\frac{1}{2^{n+k}})$$

Let $N =\bigcap_{m=1}^{\infty}(U_m)$

Show that $λ(N) = 0$.

I'm not sure how to start really:

I assume $$λ(\bigcup_{k=1}^{m}(U_k)) = λ(U_m) \leq \sum_{k=1}^{m}\sum_{n=1}^{\infty}(a_n−\frac{1}{2^{n+k}},a_n+\frac{1}{2^{n+k}})$$

by subadditivity. How do I show that $λ(N)=\bigcap_{m=1}^{\infty}(U_m) = 0$?

Please explain step by step as I am not experienced with this

• You seem to think that $\bigcup_{k=1}^m U_k=U_m$. But $U_1\supseteq U_2\supseteq U_3\supseteq \cdots$, so that's backwards. You can indeed estimate $\lambda(U_k)$ using subadditivity, but only a single sum is called for. The descending nature of the sequence $\{U_k\}$ should be helpful in computing $\lambda(N)$. – Harald Hanche-Olsen Oct 25 '16 at 15:11

## 1 Answer

For every $k\in\{1,2,\dots\}$ we have:$$0\leq \lambda\left(N\right)\leq\lambda\left(U_{k}\right)\leq\sum_{n=1}^{\infty}\lambda\left(\left(a_{n}-\frac{1}{2^{n+k}},a_{n}+\frac{1}{2^{n+k}}\right)\right)=\sum_{n=1}^{\infty}\frac{2}{2^{n+k}}=2^{1-k}$$

This can only be true if: $$\lambda\left(N\right)=0$$