TITLE: Extension of the Lebesgue measurable sets
QUESTION [17 upvotes]: 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.

REPLY [14 votes]: 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.