# $A$ a proper subset of $[0,1]$ and $m(A \cap [a,b])\leq \frac{b-a}{2}$ for every interval $[a,b]$

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.

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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 $m(A)=0$.

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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
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.

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