- #1
jostpuur
- 2,116
- 19
Assumptions: [itex]f:[a,b]\to\mathbb{R}[/itex] is some measurable function, and [itex]M[/itex] is some constant. We assume that the function has the following property:
[tex]
[x,x']\subset [a,b]\quad\implies\quad |f(x')-f(x)|\leq M(x'-x)
[/tex]
The claim: The function also has the property
[tex]
m^*(f([a,b]))\leq M(b-a)
[/tex]
I'm not sure if this is supposed to be easy, true or something else. It has been some time since I last thought about measure theory.
Some thoughts on the proof: It looks like we need the definition of the outer measure. So if we fix some [itex]\epsilon >0[/itex], then we have some intervals [itex]I_1,I_2,I_3\ldots[/itex] such that
[tex]
f([a,b])\subset \bigcup_{n=1}^{\infty} I_n
[/tex]
[tex]
\sum_{n=1}^{\infty} m(I_n) < m^*(f([a,b])) + \epsilon
[/tex]
The goal would be to prove something like
[tex]
\sum_{n=1}^{\infty} m(I_n) \lesssim M(b-a)
[/tex]
How to get there? We know
[tex]
[a,b] \subset\bigcup_{n=1}^{\infty} f^{-1}(I_n)
[/tex]
but the preimages are not intervals, so I don't see how to use this for anything.
[tex]
[x,x']\subset [a,b]\quad\implies\quad |f(x')-f(x)|\leq M(x'-x)
[/tex]
The claim: The function also has the property
[tex]
m^*(f([a,b]))\leq M(b-a)
[/tex]
I'm not sure if this is supposed to be easy, true or something else. It has been some time since I last thought about measure theory.
Some thoughts on the proof: It looks like we need the definition of the outer measure. So if we fix some [itex]\epsilon >0[/itex], then we have some intervals [itex]I_1,I_2,I_3\ldots[/itex] such that
[tex]
f([a,b])\subset \bigcup_{n=1}^{\infty} I_n
[/tex]
[tex]
\sum_{n=1}^{\infty} m(I_n) < m^*(f([a,b])) + \epsilon
[/tex]
The goal would be to prove something like
[tex]
\sum_{n=1}^{\infty} m(I_n) \lesssim M(b-a)
[/tex]
How to get there? We know
[tex]
[a,b] \subset\bigcup_{n=1}^{\infty} f^{-1}(I_n)
[/tex]
but the preimages are not intervals, so I don't see how to use this for anything.
Last edited: