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My question is the following :

is there a $\sigma$-algebra $\mathcal{T}$ (of subsets of $\mathbb{R^n}$) that contains strictly the $\sigma$-algebra $\mathcal{L}$ of Lebesgue measurable sets (in $\mathbb{R}^n$), and such that there is a measure on $\mathcal{T}$ that extends the usual Lebesgue measure on $\mathcal{L}$ ?

I guess not, but I did not find a reference.

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3 Answers 3

up vote 7 down vote accepted

It is possible to construct a translation-invariant strict extension of Lebesgue measure on $\mathbb{R}$.

Such a construction is sketched in Fremlin, Measure theory, Vol 4.I Exercise 442Yc, page 289:

Show that there is a set $A\subseteq[0,1]$, of Lebesgue outer measure $1$, such that no countable set of translates of $A$ covers any set of Lebesgue measure greater than $0$.

Hint: Let $\langle F_{\xi}\rangle_{\xi<\frak c}$ run over the uncountable closed subsets of $[0,1]$ with cofinal repetitions (4A3Fa), and enumerate the countable subsets of $\Bbb R$ as $\langle I_{\xi}\rangle_{\xi<\frak c}$. Choose inductively $x_{\xi}$, $x'_{\xi}\in F_{\xi}$ such that $x_{\xi}\notin\bigcup_{\eta,\zeta<\xi}x'_{\eta}-I_{\zeta}$, $x'_{\xi}\notin\bigcup_{\eta,\zeta\le\xi}x_{\eta}+I_{\zeta}$; set $A=\{x_{\xi}:\xi<\frak c\}$.

Show that we can extend Lebesgue measure on $\Bbb R$ to a translation-invariant measure for which $A$ is negligible.

Hint: 417A.

The Lemma in 417A reads

Let $(X,\Sigma,\mu)$ be a semi-finite measure space, and ${\cal A}\subseteq{\cal P}X$ a family of sets such that $\mu_*(\bigcup_{n\in\Bbb N}A_n)=0$ for every sequence $\langle A_n \rangle_{n\in\mathbb{N}}$ in $\cal A$. Then there is a measure $\mu'$ on $X$, extending $\mu$, such that

(i) $\mu'A$ is defined and zero for every $A\in\cal A$,

(ii) $\mu'$ is complete if $\mu$ is,

(iii) for every $F$ in the domain $\Sigma'$ of $\mu'$ there is an $E\in\Sigma$ such that $\mu'(F \mathbin{\Delta} E)=0$.}

In particular, $\mu$ and $\mu'$ have isomorphic measure algebras, so that $\mu'$ is localizable if $\mu$ is.

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In Halmos's book Measure Theory, there is a series of exercises in the chapter on extension of measures, showing that for $\sigma$-finite measures (such as Lebesgue measure), one can for each nonmeasurable set extend the measure to a measure on a $\sigma$-algebra containing that set. So the answer is yes.

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Yes. Now my answer is definitely an overkill. :-) –  Asaf Karagila Oct 8 '12 at 23:14
@Asaf It's still good to know the boundaries of what is possible. –  Michael Greinecker Oct 8 '12 at 23:18
Ok, thank you for the answer. What if we ask for the extension to also be translation invariant ? –  CyrusK Oct 8 '12 at 23:19
I guess you are refering to the Haar theorem, which, as far as I understand, says that there is a unique measure on a fixed $\sigma$-algebra (namely the Borel $\sigma$-algebra). So I fail to see how this result implies uniqueness of the $\sigma$-algebra of Lebesgue measurable sets (which is what we are looking for here). –  CyrusK Oct 8 '12 at 23:29
@CyrusK: Oh, right. I forgot about the Borel sets requirement. I'll delete my previous comment... –  Asaf Karagila Oct 9 '12 at 0:16

While definitely an overkill, let me give a partial answer.

In set theory there is a concept known as large cardinals, these are assumptions which cannot be proved from the usual axioms of ZFC, and often prove the consistency of ZFC (and therefore make a stronger theory).

One particular large cardinal axiom is the existence of a measure extending the Lebesgue measure, encompassing all the subsets of $\mathbb R$. Of course it will not be translation invariant on every set of real numbers, but it does extend the Lebesgue measure as requested.

Added after Michael's answer: combined, the answers show that we can prove "slight" increments in measurable sets; but we cannot prove that such extension will measure all sets. Not without additional axioms, anyway.

Of course I must add that if one decides to throw away the axiom of choice it is possible that the every set of reals is already Lebesgue measurable, in which case it will be impossible to extend the measure.

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