# Quick Chebychev Inequality Question

1. May 23, 2013

### Gooolati

Hello all,

I am currently working through a proof in my Real Analysis book, by Royden/Fitzpatrick and I'm confused on a part.

if f is a measurable function on E, f is integrable over E, and A is a measurable subset of E with measure less than δ, then ∫|f| < ε
A

Proof: for c>0

∫f = ∫f + ∫f <= (c)(m(A)) + 1/c ∫f
A {x in A s.t. f(x)< c} {x in A s.t. f(x)>=c} E

I understand why the first integral is less than (c)(m(A)) but I don't understand the second part.

Chebychev's inequality says that

if f is a non-negative measurable function on E then for any λ > 0

m{x in E s.t. f(x) >= λ} <= (1/λ)∫f
E

so here we would have that

m{x in A s.t. f(x)>=c} <= (1/c)∫f
A

and I don't understand how the book went from this step to getting that

∫f <= 1/c ∫f
{x in A s.t. f(x)>=c} E

any help is appreciated...thanks!

EDIT: for some reason the integrals aren't lining up with the sets they are being integrated over, hopefully it is still readable, if not please ask

2. May 23, 2013

### Office_Shredder

Staff Emeritus
If you could tex it it would be a lot easier to read

3. May 23, 2013

### Gooolati

What happened was I split the domain of A into two parts, one where f(x) < c and one where f(x) >= c

Then I applied Chebychev's inequality to the part where f(x) >= c but I was confused as to why

\int\limits_{x in A s.t. f(x)>=c} \ <= (1/c) * \int\limits_E \

edit: don't think that worked...but the second integral is integrating over E and the first one is integrating over {x in A s.t. f(x)>=c}

4. May 23, 2013

### Office_Shredder

Staff Emeritus
Is this what you're trying for (feel free to quote this post to see how tex works)

$$\int_{x\in A, f(x)\geq c} f \leq \frac{1}{c} \int_{E} f$$

It's going to depend on the definition of A if it's actually true... I notice you mention a $\delta$ and an $\epsilon$ is there supposed to be some relationship between these numbers?

5. May 23, 2013

### micromass

Could you tell us which page or which number this theorem has?

6. May 23, 2013

### Gooolati

$$\int_{x\in A, f(x)\geq c} f \leq \frac{1}{c} \int_{E} f$$

Thanks for this !

Yes the m(A) < δ and it is saying that if you integrate over this set A (a measurable subset of E), that the value of the integral will be < ε

7. May 23, 2013

### Office_Shredder

Staff Emeritus
If there is no relationship between delta and epsilon, then you are attempting to prove that your integral has a value of zero....

8. May 23, 2013

### Gooolati

it is on page 92 of Real Analysis by Royden/Fitzpatrick, Fourth Edition. It is in Section 4.6 and it is Proposition 23.

9. May 23, 2013

### Gooolati

The proposition says that for every epsilon greater than zero, there is a delta greater than zero

sorry I should have included that

10. May 23, 2013

### Gooolati

So does anyone have any tips for me? I would appreciate it very much

11. May 24, 2013

### Office_Shredder

Staff Emeritus
I'm looking at a copy of the book and I don't see any integration being done where they split the set A up into two components in the proof of proposition 23

12. May 24, 2013

### Gooolati

Really? They are trying to bound the integral of f over A. What do they do after they split up A?

13. May 24, 2013

### Office_Shredder

Staff Emeritus
I'm confused by your posts because they never split A up, which makes it hard to answer the question. They have two different functions they're integrating over (which seems to be missing from your posts and is probably related to the confusion), but they never split up A

14. May 25, 2013

### Gooolati

So this is strange...

I looked up a copy of the book online, and the proof was entirely different than the copy I have. Something else to note, sometimes in my book when referring to previous propositions or theorems, rather than saying something like "By Theorem 25" it says "By Theorem (??)"

Could my book maybe be like an incomplete version? Maybe a copy that didn't make it to the final revision

15. May 25, 2013

### micromass

Yes, I think you have some kind of draft version of the book. The proof in my copy of Royden is also completely different than what you say in this thread.