Prove that there do not exist a Lebesgue measurable set with the following property

Question

Prove that there do not exist a Lebesgue measurable set $A$ with the following property: $A\subset \mathbb R$ such that for all $0 < a < 1$, $m(A\cap [0,a] ) = a/2$.

My attempt is the following and I'm wondering if it makes sense.

Assume by contradiction that there exists a Lebesgue measurable set $A\subset\mathbb R$ such that $m(A\cap [0,a])=a/2$ for all $a\in(0,1)$ $(\star)$. Note that for all $0<a<b<1$ we have that $m(A\cap [0,b])- m(A\cap [0,a])= m(A\cap (a,b))$. This is because for any two measurable sets $E,F$ we have that $m(E\cup F) -m(E)= m(F)- m(E\cap F)$. Using the condition $(\star)$, we have that

$$m(A\cap (a,b)) = b/2-a/2= m((a,b))/2.\quad (\bullet)$$

Now, cover the interval $[0,1/2]$ using (almost disjoint) intervals $[1/2-1/3= 1/6,1/2], [1/6-1/4,1/6],\dots$ and denote each by $I_n$ for $n=1,2,\dots$. These are intervals of length $1/(n+2)$ that only overlap at points which have measure zero. Then, by the additivity of measure we have that $$m(A\cap \cup_n I_n)=\frac{1}{2}\sum_n m(I_n).$$

Note that $\cup_n I_n= (0,1/2]$ and hence the left hand side of the above equation is $$m(A\cap \cup_n I_n)= m(A\cap [0,1/2])= 1/4.$$

Also, note that $m(I_n)= 1/(n+2)$, and so the right hand side of the equation above is $$\frac{1}{2}\sum_n m(I_n)=\infty$$ being the harmonic sum.

Hence, we get a contradiction.

Edit: As pointed out in a comment the proof above is wrong.

In a comment, it was suggested to use Lebesgue differentiation theorem. Here's the Lebesgue differentiation theorem from Folland: Suppose $f\in L^1_{loc}$, then for a.e $x$ we have $$\lim_{r\to0}\frac{1}{m(E_r)}\int_{E_r}f(y)dy= f(x)$$ for every family $\{E_r\}_{r>0}$ that shrinks nicely to $x$.

By definition, a family $\{E_r\}_{r>0}$ shrinks nicely to $x$ if $E_r\subset B(r,x)$ and there is a constant $\alpha$ independent of $r$ such that $m(E_r)>\alpha m(B(x,r))$.

Fix an $a\in (0,1)$. Let $E_r= (a-r/2,a+r/2)$ for $r>0$, to be a family that shrinks nicely to $a$. Then, $m(E_r)=r$. Here, as suggested in the comment, we take $f(y)= \chi_{A\cap [0,a]}(y)$. Then, the left hand side of the theorem becomes $$\lim_{r\to0}\frac{1}{m(E_r)}\int_{E_r}f(y)dy= \lim_{r\to0}\frac{1}{r}\int_{E_r}\chi_{A\cap [0,a]}(y)dy = \lim_{r\to0}\frac{1}{r}m({A\cap [0,a]\cap E_r}). $$

We note that ${A\cap [0,a]\cap E_r}= A\cap (a-r/2,a+r/2)$ and so $m({A\cap [0,a]\cap E_r})= m((a-r/2,a+r/2))/2= r/2$ (we used the formula $(\bullet)$ derived in the wrong solution section above.) Therefore, the left hand side of the theorem is $1/2$.

The right hand side however is \chi_{A\cap [0,a]}(a)= 1$. Therefore, we get a contradiction.

Answer 1

First note that the set of all intervals $[0,a]\subseteq [0,1]$ form a $\pi$-system. Now we can use the standard result that if two probability measures agree on a $\pi$-system then they agree on the corresponding Borel $\sigma$-algebra. For concreteness, our two measures can be $\mu_1=\frac{m(X)}{2}$ and $\mu_2=m(A\cap X)$ where $m$ is the usual Lebesgue measure and $X\subseteq [0,1]$.

Now this implies that $\mu_2=\mu_1$ on all Borel sets. Where it the case that $A$ was Borel this means that $$m(A)=m(A\cap A)=\mu_2(A)=\mu_1(A)=\frac{m(A)}{2}$$

A contradiction, hence $A$ is not Borel. To show $A$ is not Lebesgue measurable use the fact that any Lebesgue measurable set differs from some Borel set via a set of measure zero.

For a reference about the $\pi$-system fact some options are

https://en.wikipedia.org/wiki/Pi-system https://www.math.lsu.edu/~sengupta/7360f09/DynkinPiLambda.pdf