If $m^*(Acap(a,b))leqfrac{b-a}{2}$. Show that $m(A)=0$.

By the definition for outer measure, there exists an open set such that $m^*(O)-m^*(A)< epsilon$ for any $epsilon > 0$ where $A subset O$.

Let $O$ be the union of disjoint intervals $(a_k,b_k)$ where $k in N$. Now, $$A= A cap O= Acap (cup (a_k,b_k))=cup (A cap (a_k,b_k))$$. So, $$m^*(A)=m^*(cup (A cap (a_k,b_k))) leq m^*(Acup (a_k,b_k)) leq Sigma(b_k-a_k)/2=m^*(O)/2$$.

Therefore, $m^*(O)/2 leq m^*(O)-m^*(A) < epsilon$ which reduces to $m^*(O) leq 2epsilon$.

Suppose that $m^*(A)>0.$ Then, there is a closed interval $I$ such that $m^*(Acap I)=r>0.$ Without loss of generality, $I=[0,1]$. Now, since $rle 1/2$, so by definition of the outer measure, $A$ must be contained in a union of intervals of length $strictly less$ than $1$. Using translation invariance of the Lebesgue measure, we may assume $Acap Isubseteq [0,delta_1]: delta_1<1$. It follows that $rle delta_1/2<1/2$ so (since $r$ is strictly less than $1/2$), there is a $0<delta_2<1/2$ such that $Acap Isubseteq [0,delta_2]$. It follows that $rle delta_2/2<delta_1/2^2$. Continuing, we may inductively define a sequence $(delta_n)$ such that $Acap Isubseteq [0,delta_n]$ and $r<delta_1/2^n$, which of course, is a contradiction, as soon as $n$ is big enough.