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In the book Real & Complex Analysis by Rudin, here states the following theorem for convolution:

Suppose that $f,g \in L^1(\mathbb R)$, then $$ h(x) := \int_{-\infty}^\infty f(x-y)g(y) dy < \infty. $$ Now, when proving this theorem, he starts with the following assertion:

"There exists Borel functions $f_0$ and $g_0$ such that $f_0 = f \text{a.e}$ and $g_0 = g \text{a.e}$."

Now, Rudin uses the definition that for the measurable space $(X, \mathcal F)$ and the topological space $(Y, \mathcal G)$, then $f:X\to Y$ is measurable if, and only if, for every open set $V$, we have $f^{-1}(V) \in \mathcal F$.

My question is then, how to prove that from a measurable function $f:\mathbb R \to \mathbb R$, I prove the existence of a Borel measurable function $f_0$ as he states in the proof. It looks to me that by the definition of Rudin, $f$ would already be Borel measurable, no? Since every open set is measurable and the open-sets generate the Borel sigma algebra.

But, the way Rudin phrases his proofs, it sounds like $f$ is not Borel measurable. How so?

Davi Barreira
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    https://math.stackexchange.com/questions/2456295/if-f-mathbbr-rightarrow-mathbbr-is-lebesgue-measurable-prove-that-the?rq=1 – blamethelag Sep 23 '21 at 18:05
  • Thanks, the question is indeed similar. But I'l be interested in understanding why my $f$ is not already Borel measurable. – Davi Barreira Sep 23 '21 at 18:16
  • Do you know what is the difference between $L(\mathbb{R})$ and $\mathcal{B}(\mathbb{R})$ ? – blamethelag Sep 23 '21 at 18:18
  • I thought that a Lebesgue Measurable function was always Borel Measurable, but it seems that's not the case. – Davi Barreira Sep 23 '21 at 18:28
  • You're mixing up $\sigma$-algebras on the domain of $f:\Bbb{R}\to\Bbb{R}$. We say $f$ is Lebesgue measurable if for every Borel set $B\subset \Bbb{R}$ (this is in the target), $f^{-1}(B)$ is Lebesgue measurable (i.e the domain). On the other hand, $f$ being a Borel function means for every Borel $B\subset \Bbb{R}$ (i.e the target) we require $f^{-1}(B)$ to be Borel. Since $\text{Borel $\sigma$-algebra of $\Bbb{R}$}\subsetneq \text{Lebesgue $\sigma$-algebra of $\Bbb{R}$}$, it follows that Borel measurable $\implies$ Lebesgue measurable. The converse (which you're asking about) is false. – peek-a-boo Sep 23 '21 at 18:36
  • Thanks, @peek-a-boo. Do you know a counter-example? – Davi Barreira Sep 23 '21 at 18:45
  • @peek-a-boo I think the converse if false only assuming the axiom of choice, right? So any counterexample uses it. – LL 3.14 Sep 23 '21 at 18:46
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    Take a Lebesgue measurable set $A\subset \Bbb{R}$ which is not Borel measurable (the fact that such a set exists follows from a cardinality argument, and is explained in Rudin early on). Then, the characteristic function $\chi_A$ is Lebesgue, but not Borel measurable (because the preimage of the singleton Borel set ${1}$ is $A$). – peek-a-boo Sep 23 '21 at 18:46
  • @LL3.14 pretty sure axiom of choice is used to exhibit a non Lebesgue-measurable set (i.e that the Lebesgue $\sigma$-algebra is not the whole power set of $\Bbb{R}$). But here I'm just speaking about something which is Lebesgue, but not Borel-measurable. But again, I'm a complete set theory novice, so maybe at some point implicitly the axiom of choice might have been used... I dont know in that case. – peek-a-boo Sep 23 '21 at 18:47
  • Grrreeat! Thanks. Things clicked :D – Davi Barreira Sep 23 '21 at 18:49
  • To prove the existence of this Borel function, would it suffice to use the fact that $C_c$ is dense in $L^1$? Then, taking a $f_0 = \lim f_n$ where $f_n \to f$ a.e? – Davi Barreira Sep 23 '21 at 19:10
  • @OliverDiaz, the proofs in the link seem a bit convoluted. Would it suffice to use the fact that $C_c$ is dense in $L^1$ and then make $f_0 = \lim \sup f_n$ with $f_n \to f$ a.e? – Davi Barreira Sep 23 '21 at 19:29
  • @DaviBarreira: the second proof is clear in my opinion (a two liner) The results easy for simple functions since as you may already now, for any Lebesgue integrable set $A$, there is a Borel set $B$ such that $A\subset B$ and $\mu(B)=\mu^*(A)$. Then by approximating functions by simple functions you get the desired result. – Mittens Sep 23 '21 at 19:32

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