Suppose $f \colon \Bbb R \rightarrow \Bbb R $ and that $f$ is increasing. Show that $f$ is measurable? Suppose $f \colon \Bbb R \rightarrow  \Bbb R $  and that $f$ is increasing. ($x$ < $x'$ $\implies $ $f(x)$ < $f(x')$). Show that $f$ is measurable. 
I am a self taught person and just started reading about this. I was wondering what this proof might look like. Can you please show it to me so I can understand what this is saying?
Does the interval $f^{-1}(a,\infty)$ work here? Is there other way to show this? 
This is the Lebesgue Measure on $  \ \Bbb R $.
I still do not get this. These hints are very vague since I just started reading about this. I only know a little from reading. If I see this proof I will be able to solve other problems like this. Thanks. 
 A: Hint: The sets $f^{-1}(a,\infty)$ are intervals for all $a$.
A: Hint: The preimage of any interval (including the empty set, singletons, rays, the whole real line) under an increasing function $\Bbb R\to\Bbb R$ is an interval (again, including the aforementioned cases).
A: Yes it is enough to check that $f^{-1}(a,\infty) \in \Sigma$ as it (usually) is the definition of measurability for a real valued function to be measurable. Where $\Sigma$ is your sigma algebra which I guess is either the Borel-sigma algebra on $\mathbb{R}$ or the Lebesgue sigma algebra on $\mathbb{R}$ since you haven't specified anything else. Now we know that all open sets are contained in the sigma algebra so pick some $x \in f^{-1}(a,\infty) = A $. We must show that there is an open ball containing $x$ that is contained in $A$. Clearly $ x > \inf f^{-1}(a,\infty)$. Hence we can find an epsilon such that $ x-\epsilon > inf \; f^{-1}(a,\infty)$. Now just take the ball $B(x,\epsilon) = \{y \in \mathbb{R} \; | \; |x-y| < \epsilon \} = (x-\epsilon, x + \epsilon) $. Now clearly every element in this set gets mapped to the interval $(a, \infty)$ by definition of epsilon and the fact that $f$ is an increasing function so $A = f^{-1}(a,\infty)$ is an open set hence it is contained in $\sigma$ thus $f$ is measurable.
