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I'm studying for my Analysis master's exam and I'm trying this question from a past exam. Here is the question:

Let $\lambda^*$ be the Lebesgue outer measure on $\mathbb R$. Suppose that $f:\mathbb R\to\mathbb R$ is a differentiable function with $|f'(x)|\leq 1$ for all $x\in\mathbb R$.
(a) Show that $|f(x)-f(y)|\leq |x-y|$ for all $x,y\in\mathbb R$.
(b) Show that for each open interval $I$ in $\mathbb R$, one has $\lambda^*(f(I))\leq \lambda^*(I)$. (Hint: recall that the outer measure of an interval is its length.)
(c) Show that for each set $A$ in $\mathbb R$, one has $\lambda^*(f(A))\leq \lambda^*(A)$.

Parts (a) and (c) are pretty straightforward -- on part (a) I know we can use the Mean Value Theorem to prove the inequality and on part (c) we can use the result in part (b) and appeal to the definition of outer measure. I'm just having trouble with part (b). I feel like we should use the result in part (a), but the issue is that we aren't guaranteed for such a function $f$ that $f((a,b))=(f(a),f(b))$. For example, the function $f(x)=\sin x$ is differentiable on $\mathbb R$ and we have $|f'(x)|=|\cos x|\leq 1$, but $f((0,\pi))=(0,1]$ and $(f(0),f(\pi))=\emptyset$. If we didn't have this issue I'd simply say $\lambda^*(f(I))=|f(a)-f(b)|\leq |a-b|=\lambda^*(I)$ by part (a), but I know that's wrong because of the counterexample displayed. Is there a way to work around this issue? Now, in the case that $f$ is increasing on $(a,b)$, we would have $f((a,b))=(f(a),f(b))$, and in the case $f$ is decreasing on $(a,b)$, $f((a,b))=(f(b),f(a))$. I'm thinking that separating a general open interval $I$ into the parts where $f$ is increasing and decreasing might be the right workaround here.

blakedylanmusic
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2 Answers2

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Fix $a<b$. It suffices to show that $f([a,b])$ has measure less than or equal to the measure of $[a,b]$. I know for continuity reasons that $f([a,b])=[c,d]$ for some $c\le d$. In particular I know $c=f(x)$ and $d=f(y)$ for some $x$ and $y$ in $[a,b]$. It follows from the MVT that $d-c\le|x-y|\le|b-a|$, so we are done.

You had the right idea; we just shouldn't focus on $f(a)$ and $f(b)$.

FShrike
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  • This makes sense, thank you! My only question is: why does it suffice to show this works on every closed interval if we want to prove this for every open interval $I$? – blakedylanmusic Aug 03 '23 at 19:54
  • Is that because $[a,b]$ and $(a,b)$ have the same outer measure? – blakedylanmusic Aug 03 '23 at 19:54
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    @blakedylanmusic Yes, $[a,b],[a,b),(a,b],(a,b)$ all have the same measure; similarly $(c,d],[c,d],[c,d),(c,d)$ all have the same measure. Finally if, say, $f((a,b))=[c,d)$ then $f([a,b])=[c,d]$ so we can use the above – FShrike Aug 03 '23 at 19:57
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The result of your problem will follow for the next Theorem:

Theorem A: If $f:(a,b)\rightarrow\mathbb{R}$ is measurable, and $f$ is differentiable on a measurable set $C\subset(a,b)$, then $$\lambda^*(f(C))\leq \int_C |f'(x)|\,dx$$

A sketch of the proof, together with some measurability considerations, can be found here

Mittens
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