Proving that any countable set of reals has Lebesgue measure zero.

[median cave]

Any countable set $A$ can have it’s elements be enumerated as $$A={x_1, x_2, x_3, \cdots}$$.

Since, we can write $A$ as
$$ A= \bigcup_i {x_i}$$

Therefore,
$$
\mu (A)= \sum_i \mu({x_i}$$

But Lebesgue measure of a singleton set is zero (because it has zero length in traditional sense), therefore, for every $i$, we have $\mu{x_i}=0$. Hence, $$\mu(A)=0$$

Why everyone else (books and Internet users) are doing it by epsilon definition? Does my proof contains some “big holes”?

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[indigo current]

Knight_Hall_Algebra

[warm wind]

Have you proved points are measurable?

And a point has measure zero?

[median cave]

it doesn't have a length

[warm wind]

But that's not a proof. Try this, each time you make a step in your proof, cite and clearly explain why. It's a good trick for these cases

'These cases' being times you're stuck

[fiery thunder]

alternatively, you can prove they have Lebesgue measure 0 without invoking anything about measures, just using the definition of outer measure for a subset of R

but yeah, as mark said that proof requires having proven the singletons have measure 0

to which, at that rate you just do the proof from the definition of outer-measure

[median cave]

I'm finding this definition of outer measure $$\mu (S) = \inf { \sum_i (b_i - a_i): S \subset \bigcup_i (a_i, b_i]}$$ very hard to grasp

Does it mean, of all the possible supersets of $S$, the infimum of their measures will be the measure of $S$?

[indigo current]

Knight_Hall_Algebra

Knight_Hall_Algebra

[vapid lake]

I think it'd be something like this

$\delta > 0, \mu (S) = \inf \qty{\sum_i (b_i - a_i) : S =\qty{p} \subset (p-\delta, p]}$

[indigo current]

a domesticated coffey

[vapid lake]

$\delta > 0, \mu (S) = \inf \qty{ p-(p-\delta) : \delta > 0}=0$

[indigo current]

a domesticated coffey

[vapid lake]

@median cave

[median cave]

i'm unaware of that definiton

[vapid lake]

i didn't use any definition

i Don't even know what Lebesgue measure is

[vapid lake]

i used what you posted :p

[median cave]

🙂

how?

[vapid lake]

here

you could even do (p-delta, p+delta)

just the fact that you could close in on the little singleton :3

[median cave]

I understand that you chose a p and created an interval from it

so that S would be contained in it

but you have written it in a very different way

[vapid lake]

not really

what have I done differently in my latex?

[devout geyser]

can you obtain a singleton from borel sets using set operations

if you know how to measure intervals, then singletons follow

and then countable sets via sigma additivity

[median cave]

I mean, we want the measure of S to be the smallest $\delta$ such that $$S \subset \left(
P -\delta/2, P+ \delta/2 \right)$$

[indigo current]

Knight_Hall_Algebra

[median cave]

P is any number in R

So, we should write

$$
\mu(S) =
\inf
{ \delta:
S \subset
(P-\delta/2, P+\delta/2) }$$

[indigo current]

Knight_Hall_Algebra

[median cave]

That’s what I was trying to claim, though informally.

That if measure is strict of the notion of length, then we must have the measure of a singleton set zero.

[fiery thunder]

Define L(I) to be the length of the interval I (intuitively what you expect)

The outer measure of A is then the 'smallest' total length
Ie find all interval covers of A, compute the sum of their lengths, taken inf

To show singletons have outer measure 0, you want to show you can find an interval cover whose length is arbitrarily small

(x0-eps,x0+eps) clearly does that for all eps>0, and the length is 2eps

So 0=<|{x0}|<2eps

[median cave]

@vapid lake Caan you see my formulation of your idea? Does it have any defects? The only problem that I have is that P is arbitrary.

[vapid lake]

that's fine

just declare what S is beforehand

[median cave]

If S= [a,b] how we would write our formulation?

[vapid lake]

just say S is a singleton bro

[median cave]

S is the interval [a,b]

[fiery thunder]

The closed interval requires invoking Heine-Borel to argue all open interval covers admit a finite subcover

then you induct on the number of intervals in the finite subcover

[median cave]

Okay

[median cave]

@fiery thunder Can you please give me an example of a set which is not Lebesgue Measurable?we are well within the boundary of $\mathbb{R}$

[indigo current]

Knight_Hall_Algebra
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[fiery thunder]

https://en.wikipedia.org/wiki/Vitali_set learn to google lol

[median cave]

To be honest, I was expecting something simpler and more straightforward from you, everywhere on the internet they are giving this same example (or a slight variation of it)

And each book that I have consulted so far, gives it as an example.

[fiery thunder]

well yeah, cause most sets are lebesque measurable

like even strange sets like continuity sets of any R to R function are measurable

all open and closed sets are measurable, continuity sets are measurable

it's honestly near impossible to find a simple non-measurable set

[median cave]

that's what I concluded myself.

They put Vitali sets only to show the relevance of the definition of Lebesgue measure.