Show that B+c is measurable If B is a borel measurable set and c is a constant, I am trying to show that B+c is measurable. B and c both are defined in R. Can any one help me on this.
 A: Define $\mathcal{F}:=\{A\subset\mathbb{R}: A+c\in \mathcal{B}(\mathbb{R})\}$
We will show that $\mathcal{F}$ is sigma-algebra.


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*$\emptyset+c=\emptyset\in \mathcal{B}(\mathbb{R})$. Hence $\emptyset\in \mathcal{F}$.

*Let $A\in \mathcal{F}$. Hence $$A+c\in \mathcal{B}(\mathbb{R})$$
$$\mathbb{R}\setminus (A+c) \in \mathcal{B}(\mathbb{R})$$
$$(\mathbb{R}\setminus A)+c \in \mathcal{B}(\mathbb{R})$$
Hence $\mathbb{R}\setminus A \in \mathcal{F}$

*Let $A_n\in \mathcal{F}, n\in \mathbb{N}$
$$A_n+c\in \mathcal{B}(\mathbb{R})$$
$$\bigcup_{n\in \mathbb{N}}(A_n+c)\in \mathcal{B}(\mathbb{R})$$
$$\bigcup_{n\in \mathbb{N}}A_n+c\in \mathcal{B}(\mathbb{R})$$
Hence $\bigcup_{n\in \mathbb{N}}A_n\in \mathcal{F}$.
Now notice that every open set belongs to $\mathcal{F}$, since translation of an open set is an open set which is borel measurable. Since $\mathcal{B}(\mathbb{R})$ is the smallest sigma algebra containing all open sets $\mathcal{B}(\mathbb{R})\subset \mathcal{F}$. Therefore for every borel measurable set $B+c \in \mathcal{B}(\mathbb{R})$.
A: By $B+c$ , i assume you mean $\{x+c : x\in B\}.$  I include open half-lines among the set of open intervals.By countable I mean not-uncountable. Let $\mu$ denote Lebesgue measure........ First, if $A$ is an open set then $A=\cup F$ where $F$ is a countable family of pairwise disjoint open intervals, and $\mu (A)=\sum_{f \in F} \mu (f). $ From this it is easy to see that $A+c$ is open and $\mu (A+c)=\mu (A).$ .... Second, if $D$ is closed and bounded, let $D\subset A$ where $A$ is a bounded open interval.We have $\mu (D+c)=\mu (A+c)-\mu ((A\backslash D)+c)=\mu (A)-\mu(A\backslash D)=\mu (D)$......Third ,if $D$ is closed and unbounded let $D=\cap_{n \in N} D_n$ where each $D_n$ is closed and bounded ,and $D_n\subset D_{n+1}$. We have $\mu (D+c)=\sup_{n \in N} \mu (D_n+c)=\sup_{n\in N}\mu(D_n)=\mu (D). $.....Finally if $B$ is measurable then for each $e>0$ let $A_e$ be open and $D_e$ closed with $D_e\subset B \subset A_e$ and $\mu (A_e \backslash D_e)<e.$ Then $D_e+c \subset B+c \subset A_e+c$ and $\mu((A_e+c)\backslash (D_e+c))=\mu (A_e\backslash D_e)<e$.......Remark: In the second and third parts,it is easy to see that if $D$ is closed then $D+c$ is closed, and that if $D$ or $D_n$ is bounded then so is $D+c$ or $D_n+c$.
A: The map $f(x) = x-c$ is continuous and so $f$ is (Borel) measurable.
Hence $f^{-1}(B)$ is Borel measurable for all Borel measurable $B$,
and since $B+ \{c\} = f^{-1} ( B)$, we have the desired result.
