# The collection of Borel sets $\mathcal{B}$ is translation invariant

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.

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

## 1 Answer

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 $$f^{-1}$$.

• 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. Apr 24, 2021 at 21:04
• Yes, it is correct but you should add that you are changing $t$ to $-t$ to get the final equality. @lorenzo Apr 24, 2021 at 23:17