If $A$ has a positive Lebesgue measure then there exist subsets which are not measurable

Question

I was thinking if there is a solution to this problem without trying to explicitly create Vitali sets in $A$. Does anyone have any ideas?

Answer 1

We do not explicitly need to produce a Vitali subset of $A$, but the existence of Vitali sets is all that we need:

Theorem. Any set $A\subseteq\mathbb R$ that is not of Lebesgue measure zero has a subset that is not measurable.

To see this, note first that if $A$ itself is nonmeasurable, we are done. Assume now that $A$ is measurable. Let $V$ be a Vitali set contained in $[0,1]$, that is, $V\subset[0,1]$ and, for any real $x$, $(x+\mathbb Q)\cap V$ is a singleton. Note that $$ A=\bigcup\{A\cap(q+V)\mid q\in\mathbb Q\}. $$ Now, either one of the sets $A\cap(q+V)$ is nonmeasurable (and we are done), or else they are all measurable, but then all of them have measure zero, and therefore so does $A$:

The point is that $A\cap(q+V)$ is measurable iff $(-q+A)\cap V$ is measurable, in which case both have the same measure, but any measurable subset of a Vitali set must be null. This is the usual argument: If $\{r_i\mid i\in\mathbb N\}$ enumerates $\mathbb Q\cap[0,1]$, and $B\subseteq V$ is measurable, then (letting $\lambda$ denote Lebesgue measure) we have that $\lambda(B+r)=\lambda(B)$ for any $r$ but also $$ \sum_i\lambda(B+r_i)=\lambda\bigl(\bigcup_i(B+r_i)\bigr)\le\lambda([0,2])=2. $$

Answer 2

I don't think the Vitali construction is useful here. Instead, I'd use the other (Bernstein's?) method of constructing a nonmeasurable set. We need two facts:

1. There are only $2^{\aleph_0}$ closed sets of real numbers.

2. Every closed set of positive measure (in fact every uncountable Borel set but we don't need that) has cardinality $2^{\aleph_0}$.

Let $A$ be a set of positive measure. Using the above facts, a straightforward transfinite induction will serve to construct two disjoint sets $B,C\subseteq A$, each of which has nonempty intersection with every closed subset of $A$ which has positive measure. (If you want to, you can just as easily construct a pairwise disjoint family of $2^{\aleph_0}$ such sets instead of only two.) It is easy to see that $B$ and $C$ are nonmeasurable. ($B$ can't have positive measure because it doesn't contain a closed set of positive measure; $A\setminus B$ can't have positive measure for the same reason. Since $A$ has positive measure, this means that $B$ is nonmeasurable.)

Answer 3

If $A$ is non-measurable, we are done. Suppose that $A$ is measurable. Then there exists $r>0$ such that $(-r, r)$ is a subset of $A-A$. We can construct a subset $N$ of $(-r, r)$ which is non-measurable. Therefore, for any $a\in A$ we have, $N+a$ is non-measurable subset of $A.$

Answer 4

Write $A= \bigcup_{n\in \mathbb N} A_n$, where $A_n = A\cap [-n,n]$.

By subadditivity, $|A|\le \sum_n |A_n|\implies |A_n|>0$ for some $n$.

Define an equivalence relation $\sim$ on $A_n$ as follows: $x\sim y\iff x-y$ is rational.

Let $\scr I$ be an index set such that $\{[x_a]: a\in \scr I\}$ is the set of equivalence classes (set by definition consists of distinct elements so the point of emphasis here is the fact that $[x_a]=[x_b]$ iff $a=b$.). Also note that $A_n= \bigcup_a[x_a]$.

Define $V=\{x_a: a\in \scr I\}$. This $V$ is a subset of $A_n$. We'll show that $V$ is not measurable via contradiction. Let $\{r_i\}_{i=1}^\infty$ be the set of all rational numbers in $[-2n-1, 2n+1]$.

Suppose on the contrary that $V$ is measurable. Let's note a few things about $V$:

$1)\,\, A_n\subset \bigcup_{i\in \mathbb N} (r_i+V)$

Pf: Take any $x\in A_n$. By the equivalence relation $\sim$, there exists an $a\in \scr I$ such that $x\in [x_a]$. It follows that $x-x_a$ is a rational no. in $[-2n, 2n]$ so must equal $r_i$ for some $i$.

$2)\,\, \{r_i+V\}_{i=1}^{\infty}$ is a disjoint collection.

Pf: $x\in (r_i+V)\bigcap (r_j+V)\implies \exists a,b\in \scr I ,$$x=r_i+x_a=r_j+x_b\implies x_a-x_b$ is rational so $[x_a]=[x_b]$. It follows that $a=b$ whence $r_i=r_j$ whence $i=j$. That is, $(r_i+V)\bigcap (r_j+V)\ne \emptyset\implies r_i=r_j.$

$3)\, r_i+V\subset [-3n-1, 3n+1]$ for all $i$.

Pf: $ x\in r_i+V\implies x=r_i+x_a$ for some $a\in \scr I$. $r_i\in [-2n,2n], x_a\in [-n,n]$ so $x\in [-3n,3n]$.

It follows by $3)$ that $\bigcup (r_i+V)\subset [-3n-1,3n+1]$. $V$ is measurable implies that all its translations are measurable so $|\bigcup_i(r_i+V)|=\sum_i |(r_i+V)|=\sum_i |V|$, where the first equality is by disjointness observed in $2)$ and the second equality is because Lebesgue measure is translation invariant.

By monotonicity in $3)$ therefore, we get $\sum_i|V|\le 6n+2\implies |V|=0$.

By $1), 2)$ however, we have $0<|A_n|\le \sum_i|V|\implies |V|>0 $.

This is a contradiction.

Answer 5

Let $A\subset\Bbb{R}$ has positive outer measure i.e $m^{\star}(A) >0$

Calim: $A$ contains a non measurable subset.

Proof: Let $ \mathcal{B}$ is a Bernstein set and $A_1=\mathcal{B}\cap A$ and $A_2=\mathcal{B^c}\cap A$

If $A_1, A_2$ both are measurable then $m^{\star}(A_1) =0=m^{\star}(A_2)$

Implies $m^{\star}(A)=0$

Hence atleast one of $A_1$ or $A_2$ is non measurable.