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

Mathematics Asked by gune on September 6, 2020

Let $$Asubsetmathbb{R}$$ such that $$m^*(Acap(a,b))leqfrac{b-a}{2}$$ for any $$a,binmathbb{R}$$. Show that $$A$$ is measurable and $$m(A)=0$$.
Here $$m^*(A)$$ is the Lebesgue outer measure of $$A$$.

In this question I’m trying to prove that $$m(A)=0$$ (So it will also imply the measurability of $$A$$).
So for that, what I was trying to do is, to write $$A$$ as a union of small intervals so that if I can get something like:
$$m^*(A)leq m^*(Abigcup_{k}(a_k,b_k))leqsumlimits_k m^*(Acap(a_k,b_k))
But I really cannot see properly how to connect the dots.

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 $$m^*(O) leq 2epsilon$$.

Answered by Umesh Shankar on September 6, 2020

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 $$0 such that $$Acap Isubseteq [0,delta_2]$$. It follows that $$rle delta_2/2. Continuing, we may inductively define a sequence $$(delta_n)$$ such that $$Acap Isubseteq [0,delta_n]$$ and $$r, which of course, is a contradiction, as soon as $$n$$ is big enough.

Answered by Matematleta on September 6, 2020