translation and dilation invariance of borel sets I am studying the book "Real Analysis" by Folland, and I have a question about the following. Folland writes on pg 37 that:

Since the collection of open intervals is invariant under translations and dilations, the same is true of Borel sets in $R$ 

I understand the "since $\dots$ dilations" part, but why does this mean that the same is true of Borel sets? It seems true, but what is the proof of this claim? 
 A: Proofs of things about $\sigma$-algebras often begin by saying "Let $A=$..." and proceed by showing that $A$ is a $\sigma$-algebra.
Here: Let $A$ be the collection of all sets $E$ such that every translate of $E$ is a Borel set and every dilate of $E$ is a Borel  set. Show that $A$ is a $\sigma$_algebra. Since open intervals are in $A$ and the Borel sets are the smallest $\sigma$-algebra containing the open intervals it follows that every Borel  set is in $A$.
A: Consider a tranlsation $\phi(x) = x+a$.  Then $\phi$ is a bijection, so it preserves countable unions, countable intersections, and complements:
$$
\phi\left(\bigcup_{n=1}^\infty A_n\right) = \bigcup_{n=1}^\infty \phi(A_n)
\\
\phi\left(\bigcap_{n=1}^\infty A_n\right) = \bigcap_{n=1}^\infty \phi(A_n)
\\
\mathbb R \setminus \phi(A) = \phi\left(\mathbb R \setminus A\right)
$$  
Can you prove this?  Do you see how to prove your result using this?
A: As David's answer describes, the primary technique to show that some property is true for all Borel sets is to show that the collection of all sets satisfying the property is a $\sigma$-algebra which contains the open sets.
With this technique in mind, we can prove the following generalization.

Let $X$ be a topological space and $f:X\to X$ be a homeomorphism. A subset $B$ of $X$ is Borel iff $f(B)$ is Borel.

To prove this, use the technique described above with the one hinted by GEdgar. That is, first use the fact that $f$ is a bijection to show that the collection of all subsets $A$ of $X$ such that $f(A)$ is Borel is a $\sigma$-algebra. Then show that this $\sigma$-algebra contains the open sets (and thus the Borel sets) by using the property that $f$ is a homeomorphism.
A: Actually, the Borel measurable function gives us the preimage of the Borel set is Borel. Thus we should use subtraction instead of addition. 
Take $B⊂R$, and $t∈\mathbb{R}$, then take $f(b) = b - t$ for all $b∈B$. This is continuous, thus Borel measurable. Therefore, $f-1(B)$ is Borel measurable. However, $f-1(B)$ is exactly $t+B$, therefore, $t+B$ is Borel measurable.
For dilation, we would use dividing instead, but just be careful when $t$ is $0$ but $0*B$ is just a single point that is open.
