Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A \subset [0,1]$ measurable and $m(A \cap [a,b])\leq \frac{b-a}{2}$ for every interval $[a,b]$. Show $m(A)=0$.

I have two ideas on how to approach this, but I don't think either of them have enough momentum to be pushed forward.

(1) First I wanted to pick a countable, disjoint set of intervals $\{I_k\}$ where $l(I_k)<\frac{b-a}{n}$ and $A \subset \cup I_k$.

(2) Another strategy I thought about was breaking $A$ into disjoint sets via $A_n := A \cap [a_n, b_n]$. However, I think this approach may be messy.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

Suppose $m(A)>0$. Note that $m(A)\leq 1$ since $A\subset [0,1]$. Let $\epsilon=m(A)/4$ which is a finite positive number. For this chosen $\epsilon>0$, we can choose a countable union of open interval $(a_i,b_i)$, $i\geq 1$ such that $A\subset \displaystyle\bigcup_{i=1}^\infty(a_i,b_i)$ and $$\tag{1}-\epsilon+\sum_{i=1}^\infty(b_i-a_i)\leq m(A)\leq \sum_{i=1}^\infty(b_i-a_i).$$ Then $$\tag{2}m(A)=m(A\cap\bigcup_{i=1}^\infty(a_i,b_i)) =m(\bigcup_{i=1}^\infty A\cap(a_i,b_i))\leq\sum_{i=1}^\infty m( A\cap(a_i,b_i))$$ where the first equality follows from $A\subset \displaystyle\bigcup_{i=1}^\infty(a_i,b_i)$, and the last inequality follows from the subadditivity of measure. On the other hand, $$\tag{3}\sum_{i=1}^\infty m( A\cap(a_i,b_i))\leq\sum_{i=1}^\infty m( A\cap[a_i,b_i]) \leq\sum_{i=1}^\infty\frac{b_i-a_i}{2}$$ where the last inequality follows from assumption.

Now combining $(1)$, $(2)$ and $(3)$, we have $$-\epsilon+\sum_{i=1}^\infty(b_i-a_i)\leq\sum_{i=1}^\infty\frac{b_i-a_i}{2}$$ which implies that (since $\sum_{i=1}^\infty(b_i-a_i)<\infty$ by $(1)$, we can do the subtraction) $$\sum_{i=1}^\infty\frac{b_i-a_i}{2}\leq \epsilon=\frac{m(A)}{4}$$ which contradicts $(1)$ when $m(A)>0$.

Therefore, we must have $m(A)=0$.

share|improve this answer
    
I have two questions. Why can we choose $\epsilon$ such that (1) is true? –  abet Oct 11 '12 at 11:03
    
This follows from definition. The measure of any set $A$ is defined to be $m(A)=\inf\{\sum_{i=1}^\infty(b_i-a_i)| A\subset\bigcup_{i=1}^\infty(a_i,b_i)\}$. –  Paul Oct 11 '12 at 11:13
    
Could I assert the finiteness of $\sum\limits_{i}(b_i-a_i)$ by this inequality: $\sum\limits_{i}(b_i-a_i)<m(A)+\epsilon$ since $A$ is finite, I can choose an epsilon such that it is true. –  abet Oct 11 '12 at 11:15
    
Yes, and we need the fact that $\sum(b_i-a_i)$ is finite in order to do the subtraction at the end. –  Paul Oct 11 '12 at 11:23
1  
well, did you think about it yourself before asking? It seems to me that you didn't. Anyway, for your last question: From (1), we have $m(A)\leq \sum_{i=1}^\infty(b_i-a_i).$ And from the last equation, we have $\sum_{i=1}^\infty\frac{b_i-a_i}{2}\leq \frac{m(A)}{4}$. Combining them, we have $\frac{m(A)}{2}\leq\sum_{i=1}^\infty\frac{b_i-a_i}{2}\leq \frac{m(A)}{4}$, that is, $\frac{m(A)}{2}\leq\frac{m(A)}{4}$. If $m(A)>0$, this is impossible. –  Paul Oct 11 '12 at 11:49

If we apply the Lebesgue differentiation theorem to the characteristic function $\chi_A$, then we have $$\chi_A(x)=\lim_{h\to 0} \frac{m(A\cap[x-h,x+h])}{2h}$$ almost everywhere, but by the properties of $A$, the function we are taking the limit of is always less than or equal to $\frac{1}{2}$. Thus, $\chi_A\leq \frac{1}{2}$ almost everywhere, but since $\chi_A$ only takes the values $0$ or $1$, it must be equal to $0$ almost everywhere.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.