Non - Measurable Sets

Does there exist a function $\mu \$ (called a measure),

0

such that

0 0,11
For all and ,
If is a countable, pairwise disjoint, Set of subsets of then

?

The answer is no! We will show this by producing a countable, pairwise disjoint Set of subsets such that

0,1 MATH

and

for all

Hence 1. and 3. cannot hold simultaneously.

The construction of

All contructions will be assumed to take place in $\QTR{Large}{[}$ 0,1 $\QTR{Large}{)}$ . In particular,

For 0 $\leq \$ $\QTR{Large}{x,y}$ $\QTR{Large}{<}$ 1,addition will be $\func{mod}$ 1

and on this Page
will denote the rational numbers in $\QTR{Large}{[}$ 0,1 $\QTR{Large}{)}$ .

Exercise: Verify that

If $\QTR{Large}{\mu }$ is translation invariant, it is also translation

invariant with respect to addition $\func{mod}$ 1.

It suffices to show that there is a Set 0,1 $\QTR{Large}{).}$

such that

for any and .

and

$\QTR{Large}{[}$ 0,1 MATH

The construction of $\QTR{Large}{C.}$

Define and equivalence relation on $\QTR{Large}{[}$ 0,1 $\QTR{Large}{)}$ by setting if

Let 0,1 $\QTR{Large}{))}$ be the Set of equivalence classes.

Invoking the Axiom of Choice on $\QTR{Large}{I}$ , Let 0,1 $\QTR{Large}{)}$ be any "Choice Set."

Specifically for any , contains a single member.

Note that

for any and

since if for and

then

in particular $\QTR{Large}{x}\$ and $\QTR{Large}{y}\$ are in the same equivalence class so $\QTR{Large}{x=y}$ .