# Thread: Deceptive Lebesgue measure theory problem

1. ## Deceptive Lebesgue measure theory problem

Let f:[0,1]->R be a measurable function and E a subset of {x : f'(x) exists}. If m(E)=0, show that m(f(E))=0.

2. ## Re: Deceptive Lebesgue measure theory problem

$\displaystyle m(f(E)) = \int_{E} f'(x) dm \leq m(E) sup_{x\in E} f'(x)=0$ for the lower bound take infimum of f'(x).

3. ## Re: Deceptive Lebesgue measure theory problem

Originally Posted by InvisibleMan
$\displaystyle m(f(E)) = \int_{E} f'(x) dm \leq m(E) sup_{x\in E} f'(x)=0$ for the lower bound take infimum of f'(x).
Where is the first equality coming from? For example, $\displaystyle m(\sin([0,\pi]))=1$, but $\displaystyle \int_{[0,\pi]}\sin'\, dm=0$.

4. ## Re: Deceptive Lebesgue measure theory problem

Well if I am not mistaken:

$\displaystyle m(\sin ([0,\pi])) = m(\sin ([0,\frac{\pi}{2}])+m(\sin ([\frac{\pi}{2},\pi]))= m([0,1])+m([1,0])=m([0,1])-m([0,1])=0$

5. ## Re: Deceptive Lebesgue measure theory problem

You're a bit confused. The image of $\displaystyle \sin([0,\pi])$ is $\displaystyle [0,1]$, which has Lebesgue measure 1. Your reversal of the endpoints of the intervals makes no sense. We're just looking at the image of the function.

6. ## Re: Deceptive Lebesgue measure theory problem

I guess by m you mean the Lebesgue measure function?

Well, it is intuitively clear that a function cannot turn a set of Lebesgure-measure zero to a set which has Lebesgue-measure non-zero.

Your example doesn't work because the sine function is differentiable on whole of [0,1].
Think about it: Is the set that you chose where sin is differentiable really a set of measure zero? And think about what you integrate over.

7. ## Re: Deceptive Lebesgue measure theory problem

Originally Posted by mastermind2007
I guess by m you mean the Lebesgue measure function?

Well, it is intuitively clear that a function cannot turn a set of Lebesgure-measure zero to a set which has Lebesgue-measure non-zero.

Your example doesn't work because the sine function is differentiable on whole of [0,1].
Think about it: Is the set that you chose where sin is differentiable really a set of measure zero? And think about what you integrate over.
That's not true. The Cantor function on $\displaystyle [0,1]$ takes a set of Lebesgue measure zero to a set of Lebesgue measure 1. It doesn't matter that $\displaystyle \sin(x)$ is differentiable everywhere. E merely needs to be a subset of of places where it is differentiable. That's besides the point, anyway. I merely was questioning the validity of the first inequality posted by InvisibleMan.

8. ## Re: Deceptive Lebesgue measure theory problem

Okay, you are right that the Cantor function does that.

I mentioned the fact that sin is differentiable everywhere because the set you chose for the sin-function isn't a set of Lebesgue-measure zero (interval [0,pi]).

As for the inequality that InvisibleMan posted:
I think there might be some absolute value signs missing.