# How measurable function spreads intervals

1. Jun 7, 2014

### jostpuur

Assumptions: $f:[a,b]\to\mathbb{R}$ is some measurable function, and $M$ is some constant. We assume that the function has the following property:

$$[x,x']\subset [a,b]\quad\implies\quad |f(x')-f(x)|\leq M(x'-x)$$

The claim: The function also has the property

$$m^*(f([a,b]))\leq M(b-a)$$

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 $\epsilon >0$, then we have some intervals $I_1,I_2,I_3\ldots$ such that

$$f([a,b])\subset \bigcup_{n=1}^{\infty} I_n$$

$$\sum_{n=1}^{\infty} m(I_n) < m^*(f([a,b])) + \epsilon$$

The goal would be to prove something like

$$\sum_{n=1}^{\infty} m(I_n) \lesssim M(b-a)$$

How to get there? We know

$$[a,b] \subset\bigcup_{n=1}^{\infty} f^{-1}(I_n)$$

but the preimages are not intervals, so I don't see how to use this for anything.

Last edited: Jun 7, 2014
2. Jun 7, 2014

### gopher_p

Would the problem be easier if you could show that the image of $[a,b]$ under $f$ was a closed interval?

3. Jun 7, 2014

### WWGD

It seems that something like f(x)=xSin(1/x) ; f(0)=0 would be a counterexample (it is continuous, so measurable), if you take some interval containing 0. I assume by m*f([a,b]) you mean the arc-length of the image.

4. Jun 7, 2014

### gopher_p

On any interval containing $0$, your function does not satisfy the primary condition of the problem at hand; just consider what the condition would imply for $f'$ where it exists.

$m^*$ usually indicates outer Lebesgue measure in this context. I'm assuming that's what it means here.

5. Jun 7, 2014

### WWGD

Sorry, I don't get your point; in any interval $[-a,a]$ , f will be bounded by a, so that , given any domain $[a,b]$, we get $|f(a)-f(b)| \leq |f(a)-f(-a)|\leq |a--(a)|=|2a|=2a$. And I don't see why f' matters, since f is only known to be measurable.

Last edited: Jun 7, 2014
6. Jun 7, 2014

### gopher_p

For any $f$ satisfying the requirement of the problem, since $|f(x+h)-f(x)|\leq M|h|$, $\left|\frac{f(x+h)-f(x)}{h}\right|\leq M$. If $f$ is differentiable at $x$ then $f'(x)\leq M$. Now your function is differentiable for all $x\neq0$, and $f'(x)=\sin(\frac{1}{x})-\frac{1}{x}\cos(\frac{1}{x})$ is unbounded near the origin. So there is no such $M$ which works for your function on any neighborhood of $0$.

7. Jun 8, 2014

### jostpuur

The counterexample by WWGD was based on misunderstanding of the problem.

I'm not sure what "arc-length of the image" means, but I was speaking about the outer measure of the image.

Yes! The outer measure $m^*(A)$ is defined by placing intervals over the set $A$, like $A\subset\bigcup_n I_n$, and making the cover optimal with some limit. The image $f([a,b])$ is defined as the set $\{y\;|\;\exists x\in[a,b]\;\textrm{s.t.}\;f(x)=y\}$. For example the length of the graph curve certainly is not the same as $m^*(f([a,b]))$, and is not very obviously related.

I just noticed that my opening post contains a redundancy in the assumptions. First it is assumed that the function is measurable, but then a new condition is introduced that makes the function continuous. If it is continuous, then of course it is measurable too. The outer measure could be replaced with the ordinary Lebesgue measure as well. I put the outer measure there to emphasize that we probably need its definition.

Last edited: Jun 8, 2014
8. Jun 8, 2014

### jostpuur

So I spoke about the definition of the outer measure as a "hint". I could have red herred myself there...

According to the assumptions the function is continuous. Therefore there exists points $x_{\textrm{min}},x_{\textrm{max}}\in [a,b]$ such that the function reaches it mimimal and maximal values there.

If $x_{\textrm{min}}<x_{\textrm{max}}$, then

$$m^*(f([a,b]))= f(x_{\textrm{max}}) - f(x_{\textrm{min}})$$
$$\leq (f(a) - f(x_{\textrm{min}})) + (f(x_{\textrm{max}}) - f(x_{\textrm{min}})) + (f(x_{\textrm{max}}) - f(b))$$
$$\leq M(x_{\textrm{min}}- a) + M(x_{\textrm{max}}-x_{\textrm{min}}) + M(b-x_{\textrm{max}}) = M(b-a)$$

If $x_{\textrm{max}}<x_{\textrm{min}}$, then

$$m^*(f([a,b]))=\cdots \leq \cdots = M(b-a)$$

similarly.

9. Jun 8, 2014

### gopher_p

It looks like you're mostly on the right track, but I'm not sure all of your equalities/inequalities are true or if they follow from your argument.

While it is true that $m^*(f([a,b]))= f(x_{\textrm{max}}) - f(x_{\textrm{min}})$, this fact is due to the fact that $f([a,b])=[f(x_{\textrm{min}}),f(x_{\textrm{max}})]$ which you have not yet demonstrated. You might find it much easier to show that $f([a,b])\subset[f(x_{\textrm{min}}),f(x_{\textrm{max}})]$, which would give you $m^*(f([a,b]))\leq f(x_{\textrm{max}}) - f(x_{\textrm{min}})$, which is good enough for this problem.

Now I'm not sure where your next inequality comes from or if it's even necessarily true. If it we me, I'd jump straight to $f(x_{\textrm{max}}) - f(x_{\textrm{min}})\leq M(x_{\textrm{max}}-x_{\textrm{min}})$ (which the key assumption here) and proceed from there.

10. Jun 8, 2014

### jostpuur

This is clearly a separate problem. Its proof comes with connectedness and components, I believe. If the image is not the interval, then it must be a union of two subsets, which are both open in the image. The preimages of open sets are open too, and contradiction follows.

It came from the fact that $f(a)-f(x_{\textrm{min}})\geq 0$ and $f(x_{\textrm{max}})-f(b)\geq 0$, so these terms could be added to the expression, only increasing its value.