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I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following exercises are Exercise 7 and Exercise 8 on p.38 in Exercises 2B in this book.

  1. Prove that the collection of Borel subsets of $\mathbb{R}$ is translation invariant. More precisely, prove that if $B\subset\mathbb{R}$ is a Borel set and $t\in\mathbb{R}$, then $t+B$ is a Borel set.
  1. Prove that the collection of Borel subsets of $\mathbb{R}$ is dilation invariant. More precisely, prove that if $B\subset\mathbb{R}$ is a Borel set and $t\in\mathbb{R}$, then $tB$ (which is defined to be $\{tb: b\in B\}$) is a Borel set.

2.40 Definition Borel measurable function
Suppose $X\subset\mathbb{R}$. A function $f:X\to\mathbb{R}$ is called Borel measurable if $f^{-1}(B)$ is a Borel set for every Borel set $B\subset\mathbb{R}$.

2.41 every continuous function is Borel measurable
Every continuous real-valued function defined on a Borel subset of $\mathbb{R}$ is a Borel measurable function.


My proof of 7:

Let $t\in\mathbb{R}$.
Let $B\subset\mathbb{R}$ be a Borel set.
Let $f:X\to\mathbb{R}$ be a function such that $X=\mathbb{R}$ and $f(x)=x-t$.
Then, $X$ is a Borel subset of $\mathbb{R}$.
Then, $f$ is a continuous real valued function defined on a Borel subset of $\mathbb{R}$.
So, by 2.41, $f$ is a Borel measurable function.
By the definition of Borel measurable function, $f^{-1}(B)$ is a Borel set for every Borel set $B\subset\mathbb{R}$.
Since $f^{-1}(B)=t+B$, $t+B$ is a Borel set.

My proof of 8:

Let $t\in\mathbb{R}$.
Let $B\subset\mathbb{R}$ be a Borel set.
If $t=0$, then $tB=\{0\}$.
Since $\{0\}$ is a closed set, $tB$ is a Borel set in this case.
Suppose $t\neq 0$.
Let $f:X\to\mathbb{R}$ be a function such that $X=\mathbb{R}$ and $f(x)=\frac{1}{t}x$.
Then, $X$ is a Borel subset of $\mathbb{R}$.
Then, $f$ is a continuous real valued function defined on a Borel subset of $\mathbb{R}$.
So, by 2.41, $f$ is a Borel measurable function.
By the definition of Borel measurable function, $f^{-1}(B)$ is a Borel set for every Borel set $B\subset\mathbb{R}$.
Since $f^{-1}(B)=tB$, $tB$ is a Borel set.


I guess my proofs are ok.
However, I cannot image what exactly Borel sets are.
For example, I know every open subset of $\mathbb{R}$ is a Borel set.
For example, I know every closed subset of $\mathbb{R}$ is a Borel set.
I know only few examples of Borel sets.
At first I did not know how to solve these exercises.
I just found 2.41 and I solved Exercises 7 and 8.
Are my proofs ok?
How can I image what exactly Borel sets are?
If someone asks me what a Borel set is, then I will answer a Borel set is an element of the collection of Borel sets.

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    $\begingroup$ I am not sure there is more easily obtained intuition other than that gained by experience. $\endgroup$
    – copper.hat
    Mar 5, 2023 at 6:08
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    $\begingroup$ @copper.hat Thank you very much for your answer. $\endgroup$
    – tchappy ha
    Mar 5, 2023 at 6:10
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    $\begingroup$ Maybe you can see them as an extension of the notion of "open" subsets. It's does not necessarily make them more intuitive, but maybe it makes you anticipate some of their properties. $\endgroup$ Mar 5, 2023 at 10:37
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    $\begingroup$ How to construct all Borel sets? Take all countable intersections of open sets. Then take all countable unions of the result. Then take all countable intesections of the result. Then take all countable unions of the result. Continue. $\endgroup$
    – A.Γ.
    Mar 5, 2023 at 11:11
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    $\begingroup$ If we accept that we cannot assign a measure to every subset of $\mathbb R$, but must restrict attention to some $\sigma$-algebra of subsets of $\mathbb R$, then the Borel $\sigma$-algebra is perhaps the first thing or the simplest thing we would think of. It’s just the $\sigma$-algebra generated by the open subsets of $\mathbb R$. $\endgroup$
    – littleO
    Mar 5, 2023 at 22:53

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I think the proofs look okay. (They could be written in a somewhat more polished fashion, but their mathematical content seems fine.)

My intuition for Borel sets on $\mathbb R$ is that they are more or less any set you can think of. All the rationals, all the irrationals, any interval or ray, any singleton, the entire line, and so on. There are sets that are not in the Borel $\sigma$-field, but they are not remotely obvious, to me.

If you ever study probability theory, you will see that working with random variables $X \colon (\Omega, \mathcal A) \to \left(\mathbb R, \mathcal B(\mathbb R)\right)$ is a very natural choice. (I believe the textbook you are studying discusses probability theory at the end.)

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  • $\begingroup$ Novice, Thank you very much for your answer. $\endgroup$
    – tchappy ha
    Mar 5, 2023 at 23:56

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