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

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• 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 $\lambda(K) > 0$.