/sci/ - Science & Math

[Lebesgue Measure No.6804678]

Let A_n be the set of reals in [0,1] whose base-2 expansion contains at most n digits:

<div class="math"> A_n = \left\{ \sum_{i=1}^n \frac{d_i}{2^i} : d_i \in \{0, 1\} \right\}</div>

Since each A_n is finite, it's Lebesgue-measurable with <span class="math"> \lambda(A_n) = 0[/spoiler].
Also, <span class="math"> A_1 \subseteq A_2 \subseteq \ldots[/spoiler].
Hence
<div class="math"> 1 = \lambda( [0,1]) = \lambda \left( \bigcup_{n=1 }^\infty A_n\right) = \lim_{n \rightarrow\infty} \lambda (A_n)</div>
<div class="math"> = \lim_{n \rightarrow\infty} 0 = 0</div>
Where am I going wrong?

[Anonymous No.6804704]

>>6804678
You proved the Lebesgue measure was 0 when n is finite and you're taking a union to when n is infinite and keeping the assertion.

At least it seems that, I don't know any math.

[Anonymous No.6804708]

>>6804704
>I don't know any math.
>Lebesgue measure was 0
>n is finite
>union
>n is infinite
kek

[Anonymous No.6804712]

>>6804708
let me try to be more formal, then

the measure of the set union of A_n from n=1 to any finite N will be 0 because of what you proved first

this doesn't imply it will in the limit to infinity

does this make any sense?

[Anonymous No.6804714]

>>6804712
i say it doesn't imply so because I know that the set of strings of {0,1} of length N is countable, and the set of strings of {0,1} of infinite length is not

[Anonymous No.6804716]

>>6804678
>Where am I going wrong?
That third equality is entirely unjustified. There isn't much more to say really.

[Anonymous No.6804717]

Pretty sure your union doesn't contain the irrationals. Consider the fractional part of sqrt(2). If it lies in the union it must be in An for some finite n, but which is it in?

[Anonymous No.6804726]

>>6804717
Dis guy got it.

[Anonymous No.6804731]

>>6804712
>I don't know any math.
><span class="math">[/spoiler]

[Anonymous No.6804737]

>>6804712
>>6804716
The third equality is a standard property of any measure.
Proof: Define <span class="math">B_i = A_i \backslash \bigcup_{j: j<i
B_j[/spoiler] where \ is set difference.
So the B_j's form a partition of <span class="math">\cup_nA_n[/spoiler] and hence
<div class="math"> \lambda \left( \bigcup_{n=1}^\infty A_n \right) = \lambda \left( \bigcup_{j=1}^\infty B_j \right)</div>
and since the B_j's are pairwise disjoint this is
<div class="math">\sum_{j=1}^\infty \lambda(B_j) = \lim_{n\rightarrow \infty} \sum_{j=1}^n \lambda (B_j) = \lim_{n\rightarrow \infty} \lambda \left(\bigcup_{j=1}^n B_j) = \lim_{n\rightarrow \infty} \lambda(A_n)</div>

>>6804717
>Pretty sure your union doesn't contain the irrationals
>If it lies in the union it must be in An for some finite n
That's it; I used an incorrect definition of set union.
Thanks anon.

[Anonymous No.6804747]

>>6804712
>>6804716
The third equality is a standard property of any measure.
Proof: Define <span class="math">B_j = A_j \backslash \left( \bigcup_{i: i<j} B_i\right)[/spoiler] where \ is set difference.
So the B_j's form a partition of <span class="math">\cup_n A_n[/spoiler] and hence
<span class="math">\lambda \left( \bigcup_{n=1}^\infty A_n \right) = \lambda \left( \bigcup_{j=1}^\infty B_j \right)[/spoiler]
and since the B_j's are pairwise disjoint this is
<div class="math">\sum_{j=1}^\infty \lambda(B_j) = \lim_{n\rightarrow \infty} \sum_{j=1}^n \lambda (B_j) =</div>
<div class="math">\lim_{n\rightarrow \infty} \lambda \left(\bigcup_{j=1}^n B_j) = \lim_{n\rightarrow \infty} \lambda(A_n)</div>

>>6804717
>Pretty sure your union doesn't contain the irrationals
>If it lies in the union it must be in An for some finite n
That's it; my definition of set union was incorrect.
Thanks anon.

[Anonymous No.6804756]

>>6804747
TeX-fixing:
<span class="math">\lim_{n\rightarrow \infty} \sum_{j=1}^n \lambda(B_j) = \lim_{n\rightarrow \infty} \lambda \left(\bigcup_{j=1}^n B_j \right) = \lim_{n\rightarrow \infty} \lambda(A_n)[/spoiler]