# Compact set and measure theory [closed]

I can´t solve this exercises

Let $\lambda$ Lebesgue measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ and $K$ compact set in $\mathbb{R}$ such that $\lambda(K)>0$. For every integer $n \geq 1$ it defines $A_n=\displaystyle \bigcup_{a \in K} \left] a-\dfrac{1}{n},a+\dfrac{1}{n}\right[$. Prove that:

a) For every $n \geq 1$,$A_{n+1} \subset A_n$.

b) $K=\displaystyle \bigcap_{n \geq 1} A_n$

c) Exists positibe integer $N$ such that $\lambda (K) > \dfrac{2}{3} \lambda (A_N)$

## closed as off-topic by Umberto P., colormegone, msteve, user147263, Brian FitzpatrickAug 11 '15 at 3:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Umberto P., colormegone, msteve, Brian Fitzpatrick
If this question can be reworded to fit the rules in the help center, please edit the question.

• what have you done ? Where is your problem ? – Surb Aug 11 '15 at 0:47

a) is obvious because the corresponding intervals get smaller. For b), clearly $K \subseteq \cap_n A_n$, so you only need to show the reverse. If $x \notin K$ then $d(x,K) > 0$ because $K$ is compact, so when $1/n < d(x,K)$ we will have $x \notin A_n$, so the reverse inclusion holds. Finally for c) note that $A_1$ is bounded because $K$ is bounded, so $\lambda(A_1) < \infty$, and hence $\lambda(K) = \lambda(\cap_n A_n) = \lim_n \lambda(A_n)$ because the $A_n$ are decreasing in $n$. By definition of limit for any $r > \lambda(K)$ we can find $N$ so large that $\lambda(A_N) < r$. Take $r=\frac{3}{2}\lambda(K)$ which is $>\lambda(K)$ since $\lambda(K) > 0$.