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:

$$m^*(A)leq m^*(Abigcup_{k}(a_k,b_k))leqsumlimits_k m^*(Acap(a_k,b_k))<epsilon$$

But I really cannot see properly how to connect the dots.

Appreciate your help

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 $0<delta_2<1/2$ 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.

Answered by Matematleta on September 6, 2020

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