Find $\lambda^* ( \bigcup _{n \geq1} [n-2^{-n}, n+2^{-n}])$

Question

Can someone show me a detailed working of this please.

Apparently it is equal to $2$.

The set evaluated would be $[0.5, 1.5] \cup [0.75, 1.25] \cup ...$

I do know that $$0 \le \lambda^* ( \bigcup _{n \geq1} [n-2^{-n}, n+2^{-n}]) \le \sum_{n=1}^{\infty}\lambda^* ( [n-2^{-n}, n+2^{-n}])=2$$

This is because the RHS would be $2/2^1+2/2^2+2/2^3+...=2$

Can't see how the actual question would be equal to $2$ as well.

This is my guess: The definition of outer Lebesgue measure is $$\lambda^*(E)=\inf \{\sum_{i \ge 1} |I_j| : I_j \, \, \text{are open intervals and} \, \, E \subseteq \bigcup_{j \ge 1} I_j \}$$

We can make our $I_j$ such as $I_1=(0.5- \varepsilon , 1.5 +\varepsilon)$, $I_2=(0.75- \varepsilon , 1.25 +\varepsilon)$ and then so on. It would be clear that $$\bigcup _{n \geq1} [n-2^{-n}, n+2^{-n}] \subseteq \bigcup_{j \ge 1} I_j $$

Then the sum of the lengths of these open intervals will be $2$ right?

Answer 1

Hint: the measure of a disjoint union of measurable sets is the sum of the measures of the sets. Have you seen a proof of this?

Edit: I noticed you forgot to shift your intervals by $n$. That could cause some trouble.