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?


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 , $\mu(f(E))=0$.

  • 2
    $\begingroup$ Monotone is not necessary; AC is enough. Try just using the definition of AC; set up your e>0 , so that there is a del>0 and cover E by small-enough intervals; smaller than del (which you can do since E has measure zero). Then you can make the cover of E as small as you want to have the total measure of f(E) be zero. $\endgroup$ – gary Nov 11 '11 at 5:31
  • $\begingroup$ I changed $f(x_k)f(x_k)$ twice, for $f(x_k),f(y_k)$ twice, which is what I think you meant. $\endgroup$ – gary Nov 11 '11 at 8:26
  • $\begingroup$ No problem, Chris, nice job. $\endgroup$ – gary Nov 11 '11 at 17:54

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 $f(E)\subseteq \cup f(w_k,z_k)$.

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    $\begingroup$ The answer does use monotone to get $f((x,y))=(f(x),f(y))$. One could handle this by taking a subinterval $(w,z)\subseteq (x,y)$ such that $f(\{w,z\})=\{\max\{f(t):t\in [x,y]\},\min\{f(t):t\in[x,y]\}$, because then $z-w\leq y-x$ and $f(x,y)=f(w,z)$ has length $|f(w)-f(z)|$. $\endgroup$ – Jonas Meyer Nov 11 '11 at 6:52
  • $\begingroup$ @JonasMeyer thank you for pointing that out, if you want you can edit it into the answer I made it community wiki because the operator basically answered his question himself. $\endgroup$ – Listing Nov 11 '11 at 7:05
  • $\begingroup$ Thanks for your expositions. $\endgroup$ – Chris Nov 16 '11 at 8:04

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