Prove that every set that is content zero is also measure zero. Prove that every set that is content zero is also measure zero.
I understand that this is true, but am not sure how exactly to prove it.
 A: Let E be a set of zero content, this means that given $\varepsilon$, there is a finite sequence of intervals $(I_n)_{n=1}^m$ such that $E\subseteq \bigcup_{n=1}^m I_n$ and the total lenght ($\sum \ell(I_n) < \varepsilon$) is less than $\varepsilon$, now pick $I_n = \emptyset$ for $n > m$, the new sequence $(I_n)_{n =1}^{\infty}$ covers $E$ and has a total lenght less than $\varepsilon$ ($\ell(\emptyset) = 0$).
A: It is sufficient to show that if $E$ is content zero, then  for each $\epsilon > 0$ there is an open set $G$ which contains $E$ and whose Lebesgue measure is less then $\epsilon$.
We consider proof for $s=1$(here $s$ is dimension of the Lebesgue measure) . Since   $E\subset R^1$ is content for $\epsilon/2 > 0$  there is a sequence of intervals $(([a_n,b_n])_{n=1}^m$ such that  $E \subseteq \cup_{n=1}^m[a_n,b_n]$ and     $\sum_{n=1}^m(b_n-a_n)<\epsilon/2$.
Let consider the family of open intervals $(]a_n-\epsilon/(8m)  ,b_n+\epsilon/(8m)[)_{n=1}^m$. It is obvious that $G:= \cup_{n=1}^m]a_n-\epsilon/(8m)  ,b_n+\epsilon/(8m)[$ covers $E$ and $l_1( \cup_{n=1}^m]a_n-\epsilon/(8m)  ,b_n+\epsilon/(8m)[)\le \sum_{n=1}^m(b_n-a_n)+m \times 2 \epsilon/(8m)\le \epsilon/2+\epsilon/4<\epsilon,$ where $l_1$ denotes one-dimensional Lebesgue measure in $R^1$.
Remark. The proof for $s>1$ is the same. We need only to show  that for each rectangle $I_n=\prod_{k=1}^s[a^{(n)}_k,b^{(n)}_k](1 \le n \le m)$ and $\epsilon/(4m)>0$  there is $t_n>0$ such that $l_s(\prod_{k=1}^s]a^{(n)}_k-t_n,b^{(n)}_k+t_n[)\le l_s(\prod_{k=1}^s[a^{(n)}_k,b^{(n)}_k[)+\epsilon/(4m)$, where $l_s$ denotes $s$-dimensional Lebesgue measure in $R^S$. We can consider a function $g_n(t)=l_s(\prod_{k=1}^s]a^{(n)}_k-t,b^{(n)}_k+t[) - l_s(\prod_{k=1}^s[a^{(n)}_k,b^{(n)}_k[)$ for $t \ge 0$. By virtue of the property of monotonous  and continuity property from above of the Lebesgue measure $l_s$ we get
 that $g(t)$ is strictly increising continuous function on $[0,=\infty[$ and $g_n(o)=0$.Hence for  $\epsilon/(4m)>0$ there is $t_n$ such that $g_n(t_n)< \epsilon/(4m)$.   
