What does it mean to "provide" a set with a $\sigma$-algebra? In the past weeks, I've often read that something is "provided" with a $\sigma$-algebra, but I don't understand what this is supposed to mean. For example, when we say that we provide $\Bbb R$ with the Borel algebra, what do we receive? A new set? And why is this even necessary?
Edit:
I would like to give some additional information on what confuses me here. As another example, take a look at one of the tasks we were given.
Let $\Omega \neq \emptyset$ be a set, $(\Xi, \Sigma)$ be a measure space and $f: \Omega \rightarrow \Sigma$ be a map. Define $\sigma(f) := \{f^{-1}(F) : F \in \Sigma \}$. 
Now, determine $\sigma(f)$ for $f: \Bbb R \rightarrow \Bbb R$, $x \rightarrow x$. In this case, let $\Bbb R$ be provided with the $\sigma$-algebra of the borel sets. 
How do I have to undertand this in this specific context? 
 A: Let me give examples that might be more familiar:


*

*Set is a set $A$ "provided" with nothing. 

*Functions are functions 
$f\colon A\to B$.


*Partially ordered set is a set $P$ "provided" with relation $<$ satisfying familiar axioms. 

*Monotone functions are functions $f\colon P\to S$ such that $x<_Py\implies f(x) <_S f(y)$, for all $x,y\in P$. 

*We might write $(P,<)$ for partially ordered set and even write $f\colon (P,<_P)\to (S,<_S)$ to denote monotone function with domain $P$ and codomain $S$.


*Group is a set $G$ "provided" with functions $m\colon G\times G\to G$, $i\colon G\to G$ and $e\colon 1\to G$ (multiplication, inverse and identity) satisfying familiar axioms.

*Homomorphisms are functions $f\colon G\to H$ such that $f(m_G(x,y)) = m_H(f(x),f(y))$, for all $x,y\in G$. 

*We might write $(G,m,i,e)$ for group, but I've never seen $f\colon (G,m_G,i_G,e_G)\to (H,m_H,i_H,e_H)$ to denote group homomorphism with domain $G$ and codomain $H$.


*Vector space over field $k$ is Abelian group $V$ "provided" with function $\pi\colon k\times V\to V$ (scalar multiplication) satisfying familiar axioms. 

*Linear maps are group homomorphisms $f\colon V\to W$ such that $f(\pi_V(\alpha,v)) = \pi_W(\alpha,f(v))$, for all $\alpha\in k$, $v\in V$. 

*We might write $(V,+_V,\pi_V)$ for vector space and even write $f\colon (V,+_V,\pi_V)\to (W,+_W,\pi_W)$ to denote linear map with domain $V$ and codomain $W$.


*Topological space is set $X$ "provided" with set $\tau\subseteq\mathcal P(X)$ (topology) satisfying familiar axioms. 

*Continuous functions are functions $f\colon X\to Y$ such that $f^{-1}(U)\in\tau_X$, for all $U\in\tau_Y$.

*We might write $(X,\tau_X)$ for topological space and even write $f\colon (X,\tau_X)\to (Y,\tau_Y)$ to denote continuous function with domain $X$ and codomain $Y$.


*Measurable space is set $X$ "provided" with set $\Sigma\subseteq\mathcal P(X)$ ($\sigma$-algebra) satisfying familiar axioms. 

*Measurable functions are functions $f\colon X\to Y$ such that $f^{-1}(A)\in\Sigma_X$, for all $A\in\Sigma_Y$. 

*We might write $(X,\tau_X)$ for measurable space and even write $f\colon (X,\Sigma_X)\to (Y,\Sigma_Y)$ to denote measurable function with domain $X$ and codomain $Y$.


As you can see there is a common theme here. We start with some set, define some structure on it and then define special class of functions that preserve said structure. 
To "provide" a set $X$ with $\sigma$-algebra means to find $\Sigma\subseteq\mathcal P(X)$ that satisfies axioms of $\sigma$-algebra. Sometimes familiar sets already have well known structures "provided" to them. For example:


*

*Set $\Bbb Z$ is Abelian group in a familiar way. We could try to "provide" $\Bbb Z$ with something exotic, but usually we don't. 

*Set $\Bbb R^n$ is a vector space over $\Bbb R$ when "provided" with familiar group structure and scalar multiplication.

*We could "provide" $\Bbb C$ with structure of real vector space if we wanted: scalar multiplication is just the usual multiplication with real numbers.

*Want a topology on $\Bbb R^n$? You don't have to look further than "providing" it with Euclidean topology.

*And what about making $\Bbb R$ into measurable space? "Provide" it with Borel $\sigma$-algebra!
Hopefully, "providing" will be clear enough by now so that we can stop using quotation marks.
If you have a function $f\colon\Bbb R\to \Bbb R$, you can provide codomain with whatever $\sigma$-algebra $\Sigma$ you want. In your case $\Sigma =\mathcal B(\Bbb R)$. Thus $\sigma(f) = \{f^{-1}(A)\mid A\in\mathcal B(\Bbb R)\}$. It might be worth to note that $\sigma(f)$ is $\sigma$-algebra, so we can provide domain with $\sigma$-algebra $\sigma(f)$. In this case, $f$ becomes a function between measurable spaces $(\Bbb R,\sigma(f))$ and $(\Bbb R,\mathcal B(\Bbb R))$. Note that $f$ didn't change in any way, its domain is still $\Bbb R$ and its codomain is still $\Bbb R$. They are just provided with (possibly) different $\sigma$-algebras. The natural question is whether $f$ is measurable function with respect to said $\sigma$-algebras. The answer is yes, and $\sigma(f)$ is the smallest $\sigma$-algebra you can provide domain $\Bbb R$ with such that $f$ is measurable (having in mind that $\sigma$-algebra on codomain is fixed).
A: We get $(\mathbb R,\mathcal B(\mathbb R))$, where $B(\mathbb R))\subset2^\mathbb R$ is the $\sigma$-algebra generated by the topology of $\mathbb R$ 
In general, the $\sigma$-algebra is a family of subsets of the given set which have the $\sigma$-algebra properties.
We need such a family of subsets because we cannot extend the natural Lebesgue measure, which measures an interval to the length of that interval $\mu([a,b])=b-a$, to the all the subsets of $\mathbb R$ (a famous result, I think due to Ulam). So our $\sigma$-algebra (in fact its completed measure) would be the maximal extension of this natural measure to subsets of $\mathbb R$
