Extension of the Lebesgue measurable sets 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. 
 A: 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.
A: 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.

A: 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.
