show that $\lim_{n\rightarrow\infty}m(V_n)=m(F)$.

Question

Let $F$ be a compact set in $R$. For each integer $n\geq1$, define

$$V_n=\bigcup_{x\in F}(x-\frac{1}{n}, x+\frac{1}{n})$$

show that $\lim_{n\rightarrow\infty}m(V_n)=m(F)$.

here $V_n$ are monotonically decreasing. I thought of using continuity from above in measure.

So, $\lim_{n}m(V_n)=m(\bigcap V_n)=m(F)$ but the concern is that, since F is not countable, I think we can not use this property. There should some other clever method using the compactness of F. Hints would be much appreciated.

Here, $m$ is the Lebesgue measure.

Answer 1

Hint: Let $A_n=\{y: d(y,F) <\frac 1 n\}$. Check that $V_n \subseteq A_n$. Note that $A_n$ decreases to $F$ so $m(V_n) \leq m(A_n) \to m(F)$.