I am trying to prove the following statement:

"Prove that the collection of Borel subsets of $\mathbb{R}$, $\mathcal{B}$, 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."

What I have done up to now:

(EDIT: I now believe only the parts in bold are necessary for the proof)

$\fbox{Let $t\in\mathbb{R}$: then $t+\mathcal{B}$ is a $\sigma$-algebra}$;

$\emptyset\in\mathcal{B}$ and $\emptyset\overset{*}{=}\emptyset +t\in t+\mathcal{B}$

*suppose for sake of contradiction that $\emptyset +t\neq\emptyset$: this means there exists an element $a\in\emptyset +t$ i.e. $a=o+t$, where $o\in\emptyset$, contradiction, since $\emptyset$ does not contain elements by definition.

$E\in t+\mathcal{B}\Rightarrow E^c=t+B^c, B^c\in\mathcal{B}$ so $E^c\in t+\mathcal{B}$;

$E_1, E_2,\dots\in t+\mathcal{B}\Rightarrow E_1=t+B_1, E_2=t+B_2, \dots\Rightarrow$ $\bigcup_{k=1}^{\infty}E_k=\bigcup_{k=1}^{\infty}(t+B_k)=t+\bigcup_{k=1}^{\infty}B_k\in t+\mathcal{B}$, since $\bigcup_{k=1}^{\infty}B_k\in\mathcal{B}$.

Now, let $O$ be an open subsets of $\mathbb{R}$: then it can be written as the countable union of open intervals in $\mathbb{R}, O=\bigcup_{k=1}^{\infty}(a_k,b_k)\in\mathcal{B}$ and if we set $t\in\mathbb{R}$ we can also write $O=t+\bigcup_{k=1}^{\infty}(a_k-t,b_k+t)\in t+\mathcal{B}$. So, $t+\mathcal{B}$ contains every open subset of $\mathbb{R}$ and since by definition $\mathcal{B}$ is the smallest $\sigma$-algebra containing all the open subsets of $\mathbb{R}$ it follows that $\mathcal{B}\subset t+\mathcal{B}.$

Let $t\in\mathbb{R}$ and and consider the function $f:\mathbb{R}\to\mathbb{R}, f(x):=x-t$: then $f$, being a continuous function, is also a Borel-measurable function so $f^{-1}(B)=B+t$ is a Borel set for every Borel set $B$ so we also have that $t+\mathcal{B}\subset \mathcal{B}$ thus $t+\mathcal{B}=\mathcal{B}$, as desired.

  • 1
    $\begingroup$ Note $\mathcal{B}=(t+\mathcal{B})+(-t)$ and the same argument gives $t+\mathcal{B}\subseteq(t+\mathcal{B})+(-t)=\mathcal{B}$. $\endgroup$ Apr 24, 2021 at 12:20

1 Answer 1


If $f$ is any homeomorphism from $\mathbb R$ to $\mathbb R$ then $f(\mathcal B)=\mathcal B$. [$f(x)=x+t$ defines a homeomorphism].

Proof: Conisder $\{A \in \mathcal B: f(A) \in \mathcal B\}$. Verify that this is a sigma algebra which contains all open sets. Hence it contains all Borel sets. This proves that $f(\mathcal B)\subseteq \mathcal B$. The reverse inclusion follows by applying the same argument to $f^{-1}$.

  • $\begingroup$ thank you for your interest in my question; I have tried to incorporate your hint into my question and I would be grateful if you could take a look at it now. Thanks. $\endgroup$
    – lorenzo
    Apr 24, 2021 at 21:04
  • 1
    $\begingroup$ Yes, it is correct but you should add that you are changing $t$ to $-t$ to get the final equality. @lorenzo $\endgroup$ Apr 24, 2021 at 23:17

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