Show that If $B\subset R$ be a Borel set then $t+B$ also be Borel set. for any $t\in R$. where R is real numbers

Question

Show that If $B\subset R$ be a Borel set then $t+B$ also be Borel set. for any $t\in R$. where R is real numbers.

I define $A= \{t+B\ is\ borel \ set\ such\ that\ t\in R \}$

Goal- Show that A is a sigma algebra contaning open sets.

First- I need show A is a $\sigma$ algebra.

Is this method correct?

Answer 1

Yes, your method is correct. Here it is in more details:

Show that if $B\subseteq \Bbb R$ is a Borel set then $t+B$ also is a Borel set, for any $t\in \Bbb R$. where $\Bbb R$ is the set of real numbers.

Proof: Let $\mathcal{B}$ be the Borel $\sigma$-algebra of $\Bbb R$. Given a fixed $t \in \Bbb R$, let
$$A_t= \{B \subseteq \Bbb R : t+B \in \mathcal{B} \}$$

Let us prove that $A_t$ is a $\sigma$-algebra.

1. $\emptyset \in A_t$. In fact, $t+\emptyset = \emptyset \in \mathcal{B}$.
2. If $B \in A_t$ then $t+B \in \mathcal{B}$. So, $ t+B^c = (t+B)^c \in \mathcal{B}$. So, $B^c \in A_t$.
3. If, for all $n \in \Bbb N$, $B_n \in A_t$ then for all $n \in \Bbb N$, $t+ B_n \in \mathcal{B}$. So $ t+ \bigcup_n B_n = \bigcup_n(t+B_n) \in \mathcal{B}$. So, $\bigcup_n B_n \in A_t$.

So, $A_t$ is a $\sigma$-algebra.

Now note that, for all open sets $O$ , $t+O$ is also an open set and thus $t+O \in \mathcal{B}$. So, for all open set $O$ , $O \in A_t$.

It follows that $\mathcal{B} \subseteq A_t$

Since this is true for any $t \in \Bbb R$, we have proved that if $B\subseteq \Bbb R$ be a Borel set then $t+B$ also be Borel set, for any $t\in \Bbb R$.

Remark 1: I answered the question as it is asked.

1. If you know that translations are continuous and that continuous functions are Borel measurable, then your question becomes trivial as a consequence of the fact that the translation by $-t$ is continuous and so Borel measurable.

2. If you know that the addition of two Borel measurable functions is a Borel measurable function, then you question also becomes trivial.

However, I suppose you are just asking for a direct proof that if $B\subseteq \Bbb R$ is a Borel set then $t+B$ also is a Borel set, for any $t\in \Bbb R$, without using any extra results.

Remark 2:

1. To prove that any continuous function $f$ is Borel measurable is to prove that $A =\{ B \subseteq \Bbb R : f^{-1}(B) \text{ is Borel } \}$ is a sigma-algebra containing all the open sets. Such proof is totally similar to the proof above in my answer.

2. To prove that, given two measurable functions $f$ and $g$, their addition $f+g$ is Borel measurable is actually a little bit trickier.

Answer 2

$t+B=f^{-1}(B)$ where $f:x\mapsto x-t$ is clearly measurable (sum of the identity map $x\mapsto x$ and the constant map $x\mapsto -t$), hence $t+B$ is a Borel set.