I have been doing this exercise and I would like to try this myself but I don't know where to begin.
The question asks to show that if $A$ is null and $f:\mathbb{R} \to \mathbb{R}$ has a continuous derivative, then $f(A)$ is $null$. Someone has already asked this question here but there are no answers in that thread and the hints written there are not helpful with my current understanding. Facts that I noticed is that if I consider $A\subseteq [0,1]$ then $f'$ must be bounded on $[0,1]$. The comment by Nicholas Still on the attached thread suggests something similar but how does the boundedness of image of the derivative of the function $f$ help when talking about measures?
To give context, we have only started reading about outer measures and Lebesgue measures and I believe this exercise only involves only those topics. Another comment in the attached thread suggests using the idea of absolute continuous functions, but I haven't come across them so far.
Any help is appreciated.
Here is the basic idea. Consider an interval $(s,t)$. $f(s,t) \subseteq [m,M]$ where $m$ and $M$ are the minimum and maximum of $f$ on $[s,t]$. If $m=f(x)$ and $M=f(y)$ then $M-m \leq C |y-x| \leq C(t-s)$ where $C$ is a bound for $|f'|$. [This follows by MVT]. Hence the measure of $f(s,t)$ does not exceed $C$ times the measure of $(s,t)$.
Now use the fact that any null set can be covered by sequence of open intervals with total length less than $\epsilon$.