# Given a differentiable function, prove that the measure of the function on null set is $0$.

Suppose $f: [0,1] \to [0,1]$ is differentiable, that its derivative $f'$ is bounded, and that $A \subset [0,1]$ is a null set. Prove that $\mu(f(A)) = 0$. I was told to use Mean Value thm. But I'm really confused since I think that $f(\emptyset) = \emptyset$, and then its measure equals to $0$. I don't know what I should really do here.

Sketch: If $[a,b]\subset [0,1],$ then $f([a,b])=[f(c),f(d)]$ for some $c,d \in [a,b].$ Now $m([f(c),f(d)]) = f(d)-f(c).$ Use the MVT on the last difference. This will show that $f$ can't enlarge the measure of an interval by more than a fixed constant factor. So when you cover a null set $E$ by a countable collection of intervals, the sum of whose measures is small, $f(E)$ will be convered by a countable collection of intervals, the sum of whose measures is small as well.

• Thank you very much! There is one problem that I have is why can we cover a null set by a countable collection of intervals. Shouldn't a null set contain nothing? I don't understand that part. – user3124 May 6 '15 at 20:06
• No, a null set in measure theory means a set of measure $0.$ – zhw. May 6 '15 at 20:08
• Oh I got it. Thank you very much! – user3124 May 6 '15 at 20:29

If $f:[0,1]\to[0,1]$ is differentiable and its derivative is bounded by some $M$, then $f$ is absolutely continuous:

Proof: Take an $\varepsilon>0$, an suppose you have a finite family of pairwise disjoint intervals $(a_n,b_n)_{n\in \mathbb{N}}$ such that $$\sum_{n=1}^{k}(b_n-a_n)<\varepsilon/M$$

Then, since for every $x,y\in [0,1]$, $x<y$, you can find $\theta\in [x,y]$ such that $f(x)-f(y)\leq f'(\theta)(x-y)\leq M(x-y)$ (MVT), you have $$\sum_{n=1}^{k}|f(b_n)-f(a_n)|\leq M\sum|b_n-a_n|<\varepsilon$$

Now, if $A\subset [0,1]$ is a null set, then for every positive $\delta$ you can find a family of intervals $(I_n)$ such that $A \subset \cup I_n$ and $\sum |I_n|<\delta$. You want to prove that $f(A)$ is a null set.

Given $\varepsilon >0$, take $I_n$ covering $A$ such that $\sum |I_n| < \varepsilon/M$. We know that $f(A)\subset \cup f(I_n)$, and $f(I_n)$ are intervals. Since $f$ is absolutely continous, we have that $$\sum |f(I_n)|<\varepsilon$$

So, you can cover $f(A)$ with intervals $f(I_n)$ that have sum of lenghts as small as you want. Then, $f(A)$ is a null set.