Dumb/Challenging conventional wisdom question possibly related to my previous question.
Why do we sometimes consider a measure space $(S, \Sigma, \mu) = (\mathbb{R}, \mathscr{B}(\mathbb{R}), \lambda)$ where $\lambda$ is Lebesgue measure rather than $(S, \Sigma, \mu) = (\mathbb{R}, \mathscr{M}(\mathbb{R}), \lambda)$ where $\mathscr{M}(\mathbb{R})$ is the set of $\lambda$-measurable subsets of $\mathbb{R}$? I mean, there are subsets of $\mathbb{R}$ that are not Borel sets but $\lambda$-measurable right? If there are none, I guess that answers the first question.
Possibly answered by above but why, in my previous question, is it 'natural' to consider $\mathscr{F}$? I'm guessing it's like why it's 'natural' to consider $\mathscr{B}(\mathbb{R})$.
Best Answer
Well i'd say it depends of the context but one reason that come to my mind is that the borel $\sigma$-algebra is simpler (and smaller) than the Lebesgue $\sigma$-algebra $\mathscr{M}(\mathbf{R})$. For a lot of things the seting of borel functions or borel sigma alegbra is enough for what you want to do, using the Lebesgue sigma algebra would only make the proofs harder or even invalidate the results you want to prove.
An example about the "harder proofs" parts : the $\sigma$-algebra $\mathscr{B}(\mathbf{R})$ is generated by the open sets of $\mathbf R$, and a lot of proofs use this fact. Unfortunatly the situation is more complex with $\mathscr{M}(\mathbf R)$.
An example about the "invalidating results" part : It's easy to show that if $f$ and $g$ are Borel then $f\circ g $ is also Borel. However, if you define a measurable function to be a function $f$ such that for every open set $U\subset \mathbf R$ you have $f^{-1}(U)\in \mathscr M (\mathbf R)$ then the composition of two measurable functions is not measurable in general.
Side note : the fact that the composition of a two measurable functions is not measurable is closely related to the fact that some functions are Borel but not Lebesgue (where $f$ is Lebesgue mean $f^{-1}(U)\in \mathscr M (\mathbf R)$ for every $U\in \mathscr M (\mathbf R))$. There is a exercice in Folland's Real analysis about that if i remember it right. But $\mathscr M (\mathbf R)$ is absolutely crucial in integration theory, indeed there are functions that are Riemann integrable but not Borel (think of the characteristic functions of some subset of the triadic cantor set).
To finish, yes $\mathscr M (\mathbf R)\backslash\mathscr B (\mathbf R)$ is nonempty. But you have the following result :
if $A\in \mathscr M (\mathbf R)\backslash\mathscr B (\mathbf R)$ then there exists two borel sets $M$ and $N$ such that $M\subset A$, $A\subset M \cup N$ and $\lambda(N)=0$ (so $A$ is a borel set up to some non Borel negligible set). Moreover one have $\lambda(A)=\lambda(M)$.