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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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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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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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You have written that if we throw axiom of choice then we can prove existence of a measure on all subsets which is laos translation invariant. Can you please provide the proof ar give some reference to study – Sushil Sep 11 '14 at 2:40
In Royden, if GCH is true then there does not exist a measure function on all subsets of R which is sigma additive and translation invariant. Can you please help me in finding proof of this? and what if we assume GCH to be false, Will there exist such a function – Sushil Sep 11 '14 at 3:45
I never wrote "prove". I said it is possible, namely it is consistent that this sort of things happen. As for the second comment, just use the axiom of choice, there are plenty of constructions of this type. Vitali sets, etc. Assuming $\sf GCH$ implies the axiom of choice. Not assuming $\sf GCH$ but assuming choice, same proofs work. – Asaf Karagila Sep 11 '14 at 12:08
Oh sorry in second comment I want to write: GCH would imply that there does not exist a measure function on all subsets of R in which length of interval is equal to measure of interval(without translation invariant) – Sushil Sep 11 '14 at 12:13
and I only want to ask(which you already told) that this statement is consistent with theory or not. And can you provide some reference for its consistency. – Sushil Sep 11 '14 at 12:15

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

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