$f:[a,b]\to[c,d]$ is absolutely continuous and bijective. Is $f^{-1}$ of bounded variation? Does $m(E) = 0$ imply $m(f(E))=0$?

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Suppose we have a function $f:[a,b]\to[c,d]$ that is absolutely continuous and bijective, where $f^{-1}:[c,d]\to [a,b]$ is the inverse. I need to determine:

  1. if $f^{-1}\in BV[c,d]$, and
  2. if it is true that $m(f(E)) = 0$ when $E\subseteq [a,b]$ with $m(E) = 0$.

Here $m$ denotes the Lebesgue measure.

I am pretty stuck on how to approach this problem. I know the definition of absolutely continuous, as well as the equivalent statements about differentiability and integration. I also know that absolutely continuous implies being of bounded variation, and that being of bounded variation is true iff the function is the difference of two bounded increasing functions.

I have all of these facts, but I'm not sure how to use them to figure out this problem. Any suggestions on how to start would be very much appreciated.

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If $f$ is continuous and one to one the it is monotonic. Hence $f^{-1}$ is also monotonic. Monotonic functions are of BV. This answers 1) (absolute continuity is not required here!). The fact that absolutely continuous functions map sets of measure 0 to sets of measure 0 is standard. This can be found in Rudin's RCA and also on MSE