Does $\sigma-$algebra of Lebesgue measurable sets of $\mathbb{R}^n$ contains non Lebesgue measurable sets?

Question

Let's denote the $\sigma-$algebra of Lebesgue measurable sets of $\mathbb{R}^n$ be $\mathcal{L}(\mathbb{R}^n)$. And $\lambda_n$ the Lebesgue measure on $\mathbb{R}^n$. We know the Borel $\sigma-$algebra of Lebesgue measurable sets of $\mathbb{R}^n$ is contained in $\mathcal{L},(\mathbb{R}^n)$, also it is generated by the collection of all $\lambda_n-$ measurable set in $\mathbb{R}^n$. My question is, is every set in $\mathcal{L}(\mathbb{R}^n)$ $\lambda_n-$ measurable? If so then this $\sigma-$algebra is exactly the collection of all $\lambda_n-$ measurable set in $\mathbb{R}^n$. But so far I didn't find any books mention this property. So is there something more? Thanks in advance.

Answer 1

The Borel $\sigma$-algebra is not generated by the collection of all $\lambda_n$-measurable sets in $\mathbb{R}^n$. The Borel $\sigma$-algebra is generated by the set of all open sets in $\mathbb{R}^n$.

The $\sigma$-algebra that is generated by the collection of all $\lambda_n$-measurable sets in $\mathbb{R}^n$ is $\mathcal{L}(\mathbb{R}^n)$, by definition of $\mathcal{L}(\mathbb{R}^n)$. So, every set in $\mathcal{L}(\mathbb{R}^n)$ is $\lambda_n$-measurable by definition.

Neither the Borel $\sigma$-algebra nor $\lambda(\mathbb{R}^n)$ contain any non-measurable sets.

Answer 2

Every set of positive Lebesgue measure has a non-measurable subset. This is basically done by showing that it contains some modified copy of the Vitali set in it.

But the $\sigma$-algebra of Lebesgue measurable sets contains no non-measurable set because by definition, its members are measurable sets. It's just the definition. The set of natural numbers contains no number that is not a natural number. So, your question is similar to the question "does the set of natural numbers $\mathbb{N}$ contain a negative integer?".

I think what is probably confusing you is the theorem that the set of Lebesgue measurable sets is a $\sigma$-algebra. In this theorem, we first consider the set of Lebesgue measurable sets and then show that it's a sigma algebra. And this is probably the source of your confusion.