8
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Michael Greinecker Apr 30 '13 at 0:18
  • $\begingroup$ 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 $I$? $\endgroup$ – PatG Apr 30 '13 at 0:40
  • $\begingroup$ 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 $\int f_+=\lim_{n\to\infty}\int f_+ 1_{[1/n,1-1/n]}$. $\endgroup$ – Michael Greinecker Apr 30 '13 at 8:45
  • $\begingroup$ 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 $0$ $\endgroup$ – leo Apr 30 '13 at 14:14
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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) $ $\endgroup$ – smiley06 May 3 '13 at 15:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.