The measure of the image of a set of measure zero Let $f$ be an absolutely continuously monotone function on $[0,1]$. Suppose $E$ has measure zero. How do I go about showing that the measure of $f(E)$ is zero?
Thanks.  
Edit: Would this work?  
Since $f$ is absolutely continuous, for every $\varepsilon \gt 0$, there is a $\delta \gt 0$ such that for a family of non-overlapping intervals $\{[x_i,y_i]\}_i$ of $[0,1]$, we have $$ \sum_i(y_i - x_i)\lt \delta ~\implies ~ \sum_i |f(y_i)-f(x_i)|\lt \varepsilon.$$
Let $(x_k,y_k)\subset [x_i,y_i]$ cover $E$. then $E\subset \bigcup (x_k,y_k)$(disjoint) and $\sum(y_k-x_k)\lt \delta.$
Also, $$f(E)\subset f\left(\bigcup (x_i,y_i)\right)=\bigcup\left(f(x_k),f(y_k)\right)~,$$
and 
$$ \mu(f(E))\leq \mu \left(\bigcup\left(f(x_k),f(y_k)\right)\right)=\sum_k \left|f(y_k)-f(x_k)\right|\lt \varepsilon.$$ 
Thus , $\mu(f(E))=0$.
 A: Your argument is basically correct.  Notice that you use monotone (and increasing) so that $f(x,y)=(f(x),f(y))$ is true for all $x<y$ (a minor technicality is that it could happen that $f(x,y)$ is a closed or half-open interval unless $f$ is strictly increasing). Other than that, there are some places that could be clearer: "Let $(x_k,y_k)\subset [x_i,y_i]$ cover $E$."   What exactly do you mean?  (What is the relationship between $k$ and $i$?  You mean to take a sequence of intervals, not just one, right?  What is $[x_i,y_i]$ here?)  In other words, the ideas are correct but could use some cleaning up and clarification, perhaps along the lines of:
Given $\varepsilon>0$, let $\delta>0$ be such that ....
Because $E$ has measure $0$, there exists a countable collection of disjoint intervals $\{(x_k,y_k)\}$ such that $E\subseteq \cup (x_k,y_k)$ and $\sum(y_k-x_k)<\delta$.  
[Perhaps consider here the technicality that some of the $(x_k,y_k)$ may not be in $[0,1]$, which can be handled by intersecting everything with $[0,1]$, possibly leaving you with one or two half-open intervals on the end, but causing no problems.]
[Insert rest of your proof.]
Monotone is not necessary; the result would be true if $f$ were only assumed absolutely continuous.  To see this, suppose the same setup with $E\subseteq \cup (x_k,y_k)$ and $\sum(y_k-x_k)<\delta$.  For each $k$, because $f$ is continuous on $[x_k,y_k]$, there exists $w_k<z_k$ in $[x_k,y_k]$ such that $\{f(w_k),f(z_k)\}=\{\min\limits_{x\in[x_k,y_k]}f(x),\max\limits_{x\in[x_k,y_k]}f(x)\}$.  Then you have $\sum(z_k-w_k)\leq\sum(y_k-x_k)<\delta$, so $\sum \mu(f(w_k,z_k))<\varepsilon$, and $f(E)\subseteq \cup f(w_k,z_k)$.
