I am reading the final chapter of baby rudin. I have trouble understanding the difference between $\mu$–measurable sets and the Borel sets. The remarks on page 309 imply that the latter is contained in the former. But from the way I see it, they are both the collection of sets, which are the final forms of open intervals after undergoing intersections, taking complements and unions. So what exactly are $\mu$–measurable sets? And why are the Borel sets $\mu$–measurable for every $\mu$?
Why is $\mu$-measurable set contained in the Borel set
borel-setsmeasure-theoryreal-analysis
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For your proof, I'm not quite sure I understand where you are going with your list of bullet points.
In the beggining it seems like you are assuming $f$ is measurable, and trying to prove that $f^{-1}(B)$ is measurable for any Borel set $B$, but your second to last bullet claims that $f^{-1}(B)$ is measurable for any borel set $B$, which then makes it seem like you are assuming this and trying to prove measurability. (But if you are assuming this, then all the other bullet points are trivial). Whichever it is you're trying to do you should make clear at the beginning of the proof.
If the first is the case (which I think it probably is) what you want to do is show that the sets form a $\sigma$-algebra which contains the open sets, as this is the reason we can say that $f^{-1}(B)$ is measurable for every Borel set.
Here is a sample proof highlighting what I've said above, since you said in the comments you would want an explicit answer.
($\Leftarrow $) if $f^{-1}(A)$ is measurable for every Borel set, then in particular $f^{-1}(A)$ is measurable for every open set, so $f$ is measurable.
($\Rightarrow$) if $f$ is measurable then we know that $f^{-1}(A)$ is measurable for every open set.
Moreover, Let $\mathcal{A}$ be the collection of sets $A$ satisfying $f^{-1}(A)$ is measurable. In particular we know that $\mathcal{A}$ contains all open sets by above. Then let $A_1, A_2, \cdots \in \mathcal{A}$
Then $f^{-1}(\bigcup_1^{\infty} A_n) = \bigcup_1^{\infty} f^{-1}(A_n)$ is measurable since it is the countable union of measurable sets, so $\bigcup_1^{\infty} A_n \in \mathcal{A}$. Similarly, $f^{-1}(A_1^{c}) = f^{-1}(A_1)^c$ is measurable since it is the compliment of a measurable set, so $A_1^{c} \in\mathcal{A}$. Thus $\mathcal{A}$ is a $\sigma$-algebra, which contain the open sets, so $\mathcal{A}$ contains all borel sets and we're done.
Is the fact that the measurable sets are a sigma-algebra equivalent to saying that unions or intersections of countable collections of measurable sets are measurable?
To be precise, the collection of measurable sets forms a $\sigma$-algebra, and measurable sets are the elements inside this $\sigma$-algebra.
If you want to prove that the collection of measurable sets forms a $\sigma$-algebra, you will show the collection contains the empty set, closed under complement and countable union (or intersections) as you said.
Edit: Given an abstract space $X$, you do not need a $\sigma$-algebra and a measure for it to become a measure space, you only need to have an algebra and a premeasure, $(X,\mathcal{A}_0, \mu_0)$. You can then extend this into a measure space $(X,\mathcal{A}, \mu)$, but the extension might not be unique if the measure is not $\sigma$-finite.
I will give a concrete example: take the set of real numbers $\mathbb{R}$.
the collection of finite union of intervals is an algebra $\mathcal{A}_0$.
a premeasure on this algebra $\mu_0$, using the fact that any finite union of intervals $\cup I_k$ can be rewritten as a finite union of disjoint intervals $\cup I'_{k'}$, define $$\mu_0 (\cup I_k) = \sum |I'_{k'}|.$$
Define an outer measure $\mu^*: 2^\mathbb{R} \rightarrow [0,\infty]$ $$\mu^*(E) = \inf \{\sum_{k=1}^\infty \mu_0(E_k) : E\subset \cup E_k \text{ and } E_k\in \mathcal{A}_0\}.$$
Using the definition of Caratheodory measurability, a set $E$ is called $\mu^*$-measurable if $$\mu^* (A) = \mu^*(A\cap E) +\mu^* (A\cap E^c)$$ for every $A\subset \mathbb{R}$.
Now we have the collection of $\mu^*$-measurable sets, one can show that this collection has the structure of a $\sigma$-algebra. When we restrict $\mu^*$ to this $\sigma$-algebra $\mathcal{A}$, we have the measure $\mu$ and the measure space $(\mathbb{R}, \mathcal{A},\mu)$.
Best Answer
For any measure $\mu$, remark (a) says that the open sets are $\mu$-measurable. A set $E$ is $\mu$-measurable simply if $\mu(E)$ is defined. Not every set is measurable though.
So the set of measurable sets $\mathfrak{M}(\mu)$ is a $\sigma$-algebra containing the open sets. Since the Borel set is the smallest $\sigma$-algebra containing the open sets, $\mathfrak{M}(\mu)$ must contain all the Borel sets. This is indeed regardless of what measure we start with.