Folland makes the following statement, "we may wish to except $\mu$-null sets from consideration in studying measurable functions. In this respect, life is a bit simpler if $\mu$ is complete":

The following implications are true iff $\mu$ is complete: if $f$ is measurable and $f=g$ $\mu$-a.e. then $g$ is measurable and if $f_n$ is measureable and $f_n \rightarrow f$ $\mu$-a.e then $f$ is measurable.

Why would we wish to exclude measure zero sets? What role is completion playing in this statement?


Consider the Borel sets on $[0,1]$ and choose $N\in [0,1]$ where $N\notin \mathscr B([0,1])$ and such that $N$ has Lebesgue measure zero. Define $f:[0,1]\to [0,1]$ by $f(x)=x.$ Now define $g:[0,1]\to [0,1]$ by

\begin{equation} g(x)= \begin{cases} f(x) & x\in [0,1]\setminus N \\ 0 & x\in N \end{cases} \end{equation}

Then, $f$ is Borel measurable, but $g$ is not, since $g^{-1}(\left \{ 0 \right \})=N.$ And yet, $f$ and $g$ differ only on a set of (Lebesgue) meaasure zero.

As another example, if you take the product measure of two measures, even if both are complete, the resulting measure is usually not complete, and you can run into problems; for example, when you prove Fubini's theorem.

Perhaps more interesting is going the other way: completing a measure can also cause difficulty. Here is an exercise in Folland:

Let $C$ be the Cantor set, $f:[0,1]\to [0,1]$ the Cantor function and take $g(x)=f(x)+x.$ Then, $g$ is strictly increasing and continuous so $h=g^{-1}$ is continuous and hence Borel measurable. But since $\lambda (g(C))=1,$ so there is an $E \subseteq g(C)$ which is not Lebesgue measurable. But $E'=g^{-1}(E)\subseteq C$ so $E'$ is Lebesgue measurable, since $\lambda(E')\le \lambda (C)=0.$

This means that $h^{-1}(E')=g(E')=E$ and so $h$ is not Lebesgue measurable.

Conclusion: working with complete measures has advantages and disadvantages.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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