Several questions about measurable functions and the sigma algebra generated by set that are symmetric around zero

Question

The following question was taken from an exam in a course I took half a year ago:

Let $\Sigma$ be the sigma-algebra on $\mathbb{R}$ that is generated from the collection $\{(-a,a)|a>0\}$.

Let $\mathcal{B}$ be the sigma-algebra of Borel on $\mathbb{R}$.

1. Is the function $x\rightarrow x^2$ measurable as a map from $(\mathbb{R},\Sigma)\rightarrow(\mathbb{R,\mathcal{B}})$?
2. Is the function $x\rightarrow x^3$ measurable as a map from $(\mathbb{R},\Sigma)\rightarrow(\mathbb{R,\mathcal{B}})$?
3. Let $f:(\mathbb{R},\mathcal{B})\rightarrow(\mathbb{R},\mathcal{B})$ be a lebesgue integrable function in all $\mathbb{R}$ i.e $f\in \mathcal{L}^1(\mathbb{R})$. Find a measurable function $g:(\mathbb{R},\Sigma)\rightarrow(\mathbb{R},\mathcal{B})$ such that: $\int_{A}gdm=\int_{A}fdm \text{ }$ for all $A\in\Sigma$.

Here $m$ is the lebesgue measure on $\mathbb{R}$.

My solution:

1. The answer is yes.

   For every $a\leq0: \{x^2\geq a\}=\mathbb{R}\in \Sigma$.

   For every $a>0: \{x^2\geq a\}=(-\infty,-\sqrt{a}] \cup [\sqrt{a},\infty)=\mathbb{R} \setminus (-\sqrt{a},\sqrt{a})\in\Sigma$.

2. No. We notice that the function $f(x)=x^3$ is not symmetric around $0$ so there is a problem. We see that:

   $f^{-1}\bigl([0,\infty)\bigl)=\{x^3\geq0\}=\{x\geq0\}=[0,\infty)\notin\Sigma$. The last holds true because $[0,\infty)$ is not symmetric around $0$ so it cannot be written as a union/intersection/complementary of symmetric sets which generates $\Sigma$.

   Thus, we found a set that is a Borel set, that its $f$ preimage is NOT in $\Sigma$. i.e. $x\rightarrow x^3$ is not measurable.

3. What I've got so far is not really helpul, but maybe it's a good direction:

   Let's assume that $f$ is measurable as a function from $f:(\mathbb{R},\Sigma) \rightarrow (\mathbb{R},\mathcal{B})$ and NOT from $(\mathbb{R},\mathcal{B})\rightarrow(\mathbb{R},\mathcal{B})$ as asked.

   We define $g(x)={|f(x)|\over2}$. We notice that for every $a\leq0: \{g\geq a\}=\{{|f|\over2}\geq a\}=\mathbb{R}\in\Sigma$.

   Now, for every $a\leq0: \{g\geq a\}=\{{|f|\over2}\geq a\}=\{f\leq-2a\}\cup\{f\geq2a\}\in\Sigma$.

   The last holds true becuase $f$ is measurable, so the sets in the last union are measurable, i.e. in $\Sigma$.

   So, $g$ is measurable. Now for the second part, we notice that for every $A\in\Sigma:$

   $\int_A{gdm}=\int_{A}{{|f|\over2}dm}={1\over2}\int_{A}{|f|dm}={1\over2}\cdot2\int_{A}{f}$. The last equality holds true because A is symmetric around $0$ and that comes from the fact that $A\in\Sigma$.

   So, we have: $\int_A{gdm}=\int_{A}{fdm}$ for all $A\in\Sigma$, as requested.

My big question is: Can we find that sort of function $g$ stated above but as a measurable function from $(\mathbb{R},\Sigma)\rightarrow(\mathbb{R},\mathcal{B})$?

Answer 1

Your solution to (3) is wrong. "The last equality holds true because $A$ is symmetric around 0": no, $A$ being symmetric has nothing to do with the equality you refer to there. You assume at the start that $f$ is $\Sigma$-measurable and not Borel measurable. Why you'd assume this is not clear, since you're asked to prove something about Borel measurable functions. In any case, it's obviously impossible - it's clear from the definitions that any $\Sigma$-measurable function is Borel measurable. The whole thing is very confused. (If you do assume that $f$ is $\Sigma$-measurable in (3) you've made it totally trivial: You can find a $\Sigma$-measurable $g$ such that $\int_A g=\int_Af$ for every $A\in\Sigma$ by letting $g=f$.)

A $\Sigma$-measurable function is precisely an even Borel function, so the natural thing to try is to let $g$ be the even part of $f$: $$g(x)=\frac{f(x)+f(-x)}2.$$

If $f$ is Borel and $g$ is as above then it's not hard to show that $g$ is $\Sigma$-measurable and $\int_A g=\int_A f$ for every $A\in\Sigma$, as required.