# Deceptive Lebesgue measure theory problem

• Jul 19th 2012, 10:35 PM
Jeh
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.
• Jul 21st 2012, 06:03 AM
InvisibleMan
Re: Deceptive Lebesgue measure theory problem
$m(f(E)) = \int_{E} f'(x) dm \leq m(E) sup_{x\in E} f'(x)=0$ for the lower bound take infimum of f'(x).
• Jul 21st 2012, 06:55 AM
Jeh
Re: Deceptive Lebesgue measure theory problem
Quote:

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

Where is the first equality coming from? For example, $m(\sin([0,\pi]))=1$, but $\int_{[0,\pi]}\sin'\, dm=0$.
• Jul 21st 2012, 08:40 PM
InvisibleMan
Re: Deceptive Lebesgue measure theory problem
Well if I am not mistaken:

$m(\sin ([0,\pi])) = m(\sin ([0,\frac{\pi}{2}])+m(\sin ([\frac{\pi}{2},\pi]))= m([0,1])+m([1,0])=m([0,1])-m([0,1])=0$
• Jul 21st 2012, 09:03 PM
Jeh
Re: Deceptive Lebesgue measure theory problem
You're a bit confused. The image of $\sin([0,\pi])$ is $[0,1]$, 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.
• Jul 27th 2012, 02:24 PM
mastermind2007
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.
• Jul 27th 2012, 03:08 PM
Jeh
Re: Deceptive Lebesgue measure theory problem
Quote:

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 $[0,1]$ takes a set of Lebesgue measure zero to a set of Lebesgue measure 1. It doesn't matter that $\sin(x)$ 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.
• Jul 28th 2012, 12:42 AM
mastermind2007
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.