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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.

PatG
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    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 00:18
  • 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 00:40
  • 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 08:45
  • 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

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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.

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smiley06
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  • 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 03 '13 at 15:23