If $f:[a,b]\to\mathbb{R}$ is increasing, does it maps Borel sets to measurable sets? Suppose $f:[a,b]\to\mathbb{R}$ is strictly increasing and left-continuous. Does it follow that $f$ maps Borel subsets of $[a,b]$ to Lebesgue measurable subsets of $\mathbb{R}$?
My intuition tells me that this is the case because the fact that $f$ is non-decreasing on a closed interval $[a,b]$ implies that $f$ has at most countably many discontinuities. But I cannot continue from there.
 A: The claim will be true if we can show that $f^{-1}$ is Borel-measurable. The idea is as follows:


*

*Find a continuous non-decreasing function $g : \Bbb{R} \to [a, b]$ extending $f^{-1}$.

*Show that $E = f([a, b])$ is Borel-measurable.


Then $g|_E = f^{-1}$ is automatically Borel-measurable, since for any Borel set $B \subset [a, b]$ we have
$$f(B) = (g|_E)^{-1}(B) = g^{-1}(B) \cap E.$$

To be specific,


*

*Define $g : \Bbb{R} \to [a, b]$ by
$$g(y) = \sup\{x \in [a, b] : f(x) \leq y \}, $$
with the convention that $\sup\varnothing = a$. It is easy to check that $g$ is continuous non-decreasing and is a left-inverse of $f$, that is, $g\circ f = \text{id}$.

*Using $g$, the image $E = f([a, b])$ is easily characterized by
\begin{align*}
E
&= \{ f(a) \} \cup \{ y : g(y-\epsilon) < g(y) \text{ for all } \epsilon > 0 \}.\\
&= \{ f(a) \} \cup \left( \cap_{k=1}^{\infty} \{ y : g(y-1/k) < g(y) \} \right).
\end{align*}
which is obviously Borel-measurable.
A: Yes, it does.
Let me just sketch the proof.  First, show that $f([a,b])$ equals $[f(a), f(b)]$ with a countable number of half-open intervals removed (corresponding to the jumps of $f$).  In particular, $f([a,b])$ is Borel.
Next, show that for any interval $[c,d] \subset [a,b]$, we have $f([c,d]) = [f(c), f(d)] \cap f([a,b])$ which is also Borel.
Now let $\mathcal{F}$ be the collection of all sets $A \subset [a,b]$ such that $f(A)$ is Borel.  We have just shown $\mathcal{F}$ contains all the closed intervals.  Since $f\left(\bigcup_n A_n\right) = \bigcup_n f(A_n)$, $\mathcal{F}$ is closed under countable unions.  And since $f$ is injective, we have $f(A^c) = f([a,b]) \setminus f(A)$ so that $\mathcal{F}$ is closed under complements.  Thus $\mathcal{F}$ is a $\sigma$-algebra and it contains all the Borel subsets of $[a,b]$.
A: The answer is YES.
A more elementary proof. If $f$ is left continuous and strictly increasing, then 
$$
f\big[[a,b]\big]=\big(f(a),f(b)\big)\setminus\bigcup_{i\in I}(c_i,d_i],
$$
where, the intervals $(c_i,d_i]$ are disjoint and countable.
In order to show that $f$, which is also 1-1, takes Borel sets to Borel sets, it suffices to show that $f\big[(A,B)\big]$ is Borel, for every $A,B\in\mathbb R$. But, also clearly
$$
f\big[(A,B)\big]=\Big(\big(f(a),f(b)\big)\setminus\bigcup_{i\in I}(c_i,d_i]\Big)\cap \big(f(A),f(B)\big),
$$
which is indeed a Borel set!
