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.

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.

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.