The h Bar

@ACuriousMind My definition of "set of measure zero" is one for which, given $\epsilon >0$ there exist a collection of open rectangles $R_i$ such that $\sum V(R_i)<\epsilon$, where $V(R_i)$ is the product of side lengths (volume)

2 hours ago, by 0celo7

I'm trying to prove that one can substitute rectangles with cubes as follows:

And that's where I fail

yuggib

1:23 AM

@0celo7 your definition of a set of measure zero is not the usual one

0celo7

@yuggib It's the usual one in geometry.

yuggib

I doubt it, anyways go on

0celo7

Why do you doubt it

yuggib

because measure theory is the same in analysis and geometry

0celo7

I have 4 texts on geometry here that use it.

@yuggib According to two of them it's equivalent to the "usual" definition.

But this is easier to work with because you don't need measure theory explicitly

yuggib

1:26 AM

ok, so you have to tell me what the space is, what rectangles are

0celo7

@yuggib space is $\Bbb R^n$

yuggib

and still in my opinion is much worse to work with ;-P

0celo7

rectangles are sets of the form $(0,a_1)\times\cdots\times(0,a_n)$

squares are sets of the form $(0,a_1)\times\cdots\times(0,a_1)$

I've moved them to the origin here for ease of writing

yuggib

ok, but I have to say that this is really not the way you should define a set of measure zero

0celo7

According to a hint in a book, one must use the fact that any rectangle can be covered by cubes with total volume no greater than $2^n\cdot$volume of the rectangle

yuggib

1:28 AM

for example because it defines only sets of Lebesgue measure zero

0celo7

Yeah, so?

I'm only interested in that

yuggib

it's just very misleading notation

you want to prove what exactly?

that you can substitute rectangles with squares in the definition

?

0celo7

Yes, I can do it once I have proven the hint

I've completed the proof from there

yuggib

in two dimensions you can prove it?

0celo7

Actually in two dimensions I was able to get it down to $2\cdot$vol(R)

not $4$

you can read my proof above somewhere

@yuggib chat.stackexchange.com/transcript/message/30391814#30391814 ff

@yuggib My proof in $n$ dimensions might work like this:

Pick some nice number $a$ less than each of the side lengths.

Partition off each ...vertex?

yuggib

1:36 AM

edge

0celo7

What are the thingies called

Right, I knew the German word for it :P

So partition each edge as follows:

(0,a), (a/2,3a/2), (a,2a), ...

Then you go all the way until you run over

This will be $2\lceil a_i/a\rceil-1$ paritions

where we are talking about the $i$th edge

So you do this for each edge

yuggib

hint: take $a$ to be the minimum of the edge lengths

0celo7

@yuggib If you read above I already did that.

>:(

But then I got $2^{n-1}$

But then you just take products of all of these intervals and get the squares

yuggib

$2^{n-1}\leq 2^n$, so where is the problem?

0celo7

WLOG let $a_1$ be the shortest one

So you end up with:

21 mins ago, by 0celo7

We can cover the rectangle with $$\prod_{i=2}^n 2\left(\lceil\frac{a_i}{a_1}\rceil -1\right)$$ squares

Now, the ceiling of $a_i/a$ is bounded by $a_i/a+1$, so the product is bounded by $\prod 2a_i/a_1$

Then multiply that by the volume of the cube, $a_1^n$ and you get a bound of $2^{n-1}a_1\cdots a_n$?

@yuggib because the hint said $2^n$

So I'm not convinced this is correct.

yuggib

1:49 AM

@0celo7 I am busy now, I will come back to you later

;-)

0celo7

>:(

Fine I will type up a nice proof

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

I'm sure the good professor would prefer that ;-P

0celo7

2:08 AM

wait it's wrong

I misplaced parentheses

yeah this is wrong

I also flipped an inequality.

err...maybe it does work out

are you still there?

working on the proof

I see

I think it works, just typing it out

yuggib

2:23 AM

anyways, it seems that sets with your definition are called null sets

0celo7

hmm

crap infinite sum fun

I need to work on the last part.

yuggib

you have to replace some $a_n$ with $a_1$ in denominators

0celo7

oops

yuggib

at the beginning

0celo7

yeah

Wow I just deleted a song from YG's new album

Title: "Fdt" Hook: "Fuck Donald Trump"

seriously?

@yuggib did you read my proof?

yuggib

2:36 AM

the inequality is in reverse, so it's not good as it is...

0celo7

what inequality

whoops there's another typo

yuggib

ok, it seems good

0celo7

which inequality is in reverse?

yuggib

no it is not

0celo7

@yuggib I do have a question left

How exactly should I write the first sum on the last line

yuggib

2:38 AM

but are you sure you're not forgetting the cube for the edge $E_1$?

0celo7

@yuggib you mean in the product?

yuggib

if there is that one more, then you really have $2^n$

0celo7

or where

yuggib

yep

0celo7

no because $a_1/a_1=1$

So you just get a contribution of 1 to the product from $E_1$

If you work it out explicitly in 2D you can see it

yuggib

2:40 AM

yes but you're counting their number

0celo7

Try fitting cubes in e.g. the 1x3 rectangle

you get 3+2=5

Which SHOULD be $k$ as given above

yuggib

in the $1\times 3$ rectangle there are three cubes of edge $1$

0celo7

But these are open cubes

You're leaving out the bits between the cubes

so you need 2 more to cover those bits

yuggib

it is not optimal

you can use four overlapping cubes

0celo7

I don't care if it's optimal, I care if it works

Can you generalize your covering?

I'm still not sure why you think mine is wrong

yuggib

2:44 AM

I don't think it is wrong

I was just saying

0celo7

so my $k$ is indeed correct?

@yuggib but yes I do have a question

So I construct a countable set $C^\beta{}_{(\alpha)}$

How do I consolidate those indices to a single index set $M$ so that $\sum_{\mu\in M} V(C_\mu)$ makes sense

yuggib

what is the range of $\beta$ and $\alpha$?

anyways, you form couples $M=\{(\alpha,\beta),\alpha\in A,\beta\in B\}$

0celo7

$\alpha$ is some countable set and $\beta$ is finite

@yuggib so...how would you write the last line

yuggib

either you define $M$ to be as I said above, and then $\sum_{\mu\in M}V(C_\mu)$

or you put two sums everywhere

0celo7

Well I'm afraid I don't see why that works I guess

yuggib

2:50 AM

?

0celo7

Or what your $M$ really means

I'm not sure what $C_\mu$ is

yuggib

you understand what a couple $(\alpha,\beta), \alpha\in A,\beta \in B$ is, right?

0celo7

I have the countable set of rectangles $R_\alpha$ and to each $R_\alpha$ I associate a finite number of cubes $C^\beta_{(\alpha)}$

I need to get a countable set $C_\mu$ of cubes now

@yuggib probably not

yuggib

in other words, $M=A\times B$, $\times$ being the cartesian product

0celo7

I know what $\times$ is, I'm not stupid

yuggib

2:52 AM

so what is the problem with $M$?

0celo7

I don't know what $\sum_M$ means

yuggib

means the sum over all possible couples $(\alpha,\beta)$, with $\alpha\in A$ and $\beta\in B$

and $C_\mu :=C^\beta_{(\alpha)}$

0celo7

I know...

yuggib

given that $\mu=(\alpha,\beta)$

you are giving contradictory statements

0celo7

I guess I'm worried about changing the order of sums

yuggib

2:54 AM

you know and don't know at the same time

0celo7

and that thingie

erm

yuggib

you're not changing it

0celo7

reordering sums

yuggib

if $B$ depends on $\alpha$

then you have to be careful

0celo7

Well...it might.

Crap

yuggib

2:55 AM

then you have to be more careful

0celo7

I guess I have to use that covering there?

Because using that specific covering I get the same number of cubes in each covering

And the inequalities still hold, obviously.

yuggib

which covering? the overlapping one?

0celo7

But my advisor is a geometric analyst, he probably won't want that level of pedantry.

@yuggib Yeah

yuggib

that's not difficult to define

0celo7

Because then $\beta$ takes the same number of values for every $\alpha$

namely $k$

yuggib

2:57 AM

I see

to make that covering of an edge, take $a_i/a_1= n+ r$, where $n$, $r$ are natural numbers with $0\leq r< a_1$

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