# Image of a set of zero measure has zero measure

I am studying for my final and got stuck on the following problem from the previous year. I put my attempt below.

Suppose that $I\subset \mathbb{R}$ is an open interval, $f:I\rightarrow \mathbb{R}$ is differentiable on $I$ and its derivative is continuous on $I$. If $a,b\in I$ and $E\subseteq [a,b]$ is of Lebesgue measure zero, show that $f(E)$ is a set of Lebesgue measure zero.

I think that since $E$ is a set of measure zero for every $\varepsilon>0$, there is a countable covering of $E$ with disjoint open intervals $E\subset \cup_{i\geq 1} U_i$ where $|U_i|=d_i$, so that $\sum_{i=1}m^\ast(U_i)=\sum_{i=1} d_i<\varepsilon$, then: $$m^\ast(f(E))\leq \sum_{i=1} m^\ast(f(U_i))\leq \sum_{i=1}|(f(u_{i1}),f(u_{i2}))|=\sum_{i=1}|f(u_{i1})-f(u_{i2})|\leq \sup_{a\leq t\leq b}|f'(t)| \varepsilon,$$ so $m^\ast(f(E))$ has measure zero as $f'$ is bounded on $[a,b]$.

Howerver, I am not sure why $\sum_{i=1} m^\ast(f(U_i))\leq \sum_{i=1}|(f(u_{i1}),f(u_{i2}))|$.

I am also trying to see why this would imply that for a set $E\subseteq I$ with measure zero, $f(E)$ is a set of measure zero.

• You are on the right track. But the cover with open intervals will in general not be a disjoint cover (the rationals have measure zero). The crucial property of $f$ that helps here is Lipschitz continuity. – Michael Greinecker Apr 30 '13 at 0:18
• So that if $f:I\rightarrow \mathbb{R}$ and there is $M>0$ s.t. $|f(x)-f(y)|\leq M|x-y|$ for all $x,y\in I$ and $E$ has measure zero, $E\subseteq \cup_{i\geq 1} U_i$ where each $U_i$ is an open interval and $\sum_{i=1}m^\ast(U_i)=\sum_{i=1}|U_i|<\varepsilon$, so by the Lipschitz condition $|f(U_i)|\leq M |U_i|$ and $m^\ast(f(E))\leq M\varepsilon$. So this shows the first part of the question since for $f:[a,b]\rightarrow \mathbb{R}$, $|f(x)-f(y)|\leq \sup_{a<t<b}|f'(t)| |x-y|\leq M |x-y|$. Does the second part follow then by extending $f$ to be differentiable on the closure of $I$? – PatG Apr 30 '13 at 0:40
• In general, you cannot extend $f$ that way. Think of $f(x)=1/x\mathrm{sin}(x)$. Split into positive and negative parts and show that both have integral zero by using the monotone convergence theorem. Suppose $I=(0,1)$ If $f_+$ is the positive part, you have $\int f_+=\lim_{n\to\infty}\int f_+ 1_{[1/n,1-1/n]}$. – Michael Greinecker Apr 30 '13 at 8:45
• Following your last comment, you are in the right track. One last advice, do not worry about the differentiability of $f$ in the boundary. Work on the interior of the interval, after all the boundary has measure $0$ – leo Apr 30 '13 at 14:14

Essentially you are taking $E\subset U$ with $m^*(U)<\epsilon$ where WLOG $U = \cup_i(a_i,b_i)$ for such disjoint intervals and then estimating $$\sum_i |f(b_i)-f(a_i)| = \sum_i \left|\int^{b_i}_{a_i}f'(t)dt\right| \leq \sum_i \int^{b_i}_{a_i}|f'(t)|dt = \int_U |f'(t)|dt \leq \epsilon \sup |f'(t)|$$ But as you rightly doubted, although $U = \cup_i(a_i,b_i)$, it doesn't necessarily imply $f(U) \subset \cup_i(f(a_i),f(b_i))$ from which you may conclude from above $$m^*(f(U))\leq \sum_i |f(b_i)-f(a_i)| \leq \epsilon \sup|f'(t)|$$ However the implication is true for monotonically increasing functions. And infact for your function can be expressed as a difference of two increasing functions, so doing the estimate for increasing functions is good enough.To see this last statement you need to define the total variation, as follows $$g(x) = V^x_a(f) = \sup\{\ \sum_i|f(x_i)-f(x_{i-1})| \ a= x_0 <x_1<...<x_n = x\}$$ Where the sup is taken over all partitions. So you can observe that for $a < x<y\leq b$ you have $$V^y_a(f) = V^x_a(f) + V^y_x(f)$$ which makes $g$ increasing, and also observe $|f(y) -f(x)| \leq V^y_x(f) = V^y_a(f)-V^x_a(f)$ and hence $g(y)-f(y) \geq g(x)-f(x)$ making $g-f$ also increasing, and $f = g - (g-f)$ and you can estimate $g(U)$ and $(g-f)(U)$ seperately.
• In this case the function is wonderful, that is why $g$ is finite in compact interval. These are pretty much finer estimates, you can also try taking $f$ Lipschitz and estimate the outer measure of $E$ by cubes, you will find $m^*(f(E)) \leq C(Lip(f))m^*(E)$ – smiley06 May 3 '13 at 15:23