Infinite Union of Non-disjoint Sets

Yagoda
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Homework Statement


To give some context, I'm trying to show that \mu(\bigcup^{\infty}_{k=1}A_{k})\leq \sum^{\infty}_{k=1}\mu(A_{k}) where μ is the Lebesgue measure and the A's are a countable set of Borel sets.

Since the A's may not be disjoint, I'm trying to rewrite the left side of the equation to somehow show that the elements that are present in more than one set are not being counted in the same way that they are on the right, which causes the left side to be smaller (or equal if all sets are disjoint).

Homework Equations





The Attempt at a Solution


In class we were given the hint to consider elements that are only in one set, present in 2 sets, in 3 sets, etc. and use this to rewrite the inequality.

The elements that are only in one set particular set A_{k} = A_{k} \setminus \bigcup^{k-1}_{i=1} A_{i}. When I union all of these, it looks pretty messy.
I'm stuck on figuring out how to write elements that are present in more than one set. Does this approach look like it's on the right track?

(From what I've read it seems to common to build this up using the outer Lebesgue measure, but we haven't covered that. It seem like the way we're going about this is kind of unconventional)
 
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use the definition of external measure as infimum.
 
Yagoda said:

Homework Statement


To give some context, I'm trying to show that \mu(\bigcup^{\infty}_{k=1}A_{k})\leq \sum^{\infty}_{k=1}\mu(A_{k}) where μ is the Lebesgue measure and the A's are a countable set of Borel sets.

Since the A's may not be disjoint, I'm trying to rewrite the left side of the equation to somehow show that the elements that are present in more than one set are not being counted in the same way that they are on the right, which causes the left side to be smaller (or equal if all sets are disjoint).

Homework Equations





The Attempt at a Solution


In class we were given the hint to consider elements that are only in one set, present in 2 sets, in 3 sets, etc. and use this to rewrite the inequality.

The elements that are only in one set particular set A_{k} = A_{k} \setminus \bigcup^{k-1}_{i=1} A_{i}. When I union all of these, it looks pretty messy.
I'm stuck on figuring out how to write elements that are present in more than one set. Does this approach look like it's on the right track?

(From what I've read it seems to common to build this up using the outer Lebesgue measure, but we haven't covered that. It seem like the way we're going about this is kind of unconventional)

Can you show \mu(A \cup B) \leq \mu(A) + \mu(B)? If so, you can get by induction that V_n \equiv \mu \left( \cup_{i=1}^n A_i \right) \leq \sum_{i=1}^n \mu(A_i) \leq \sum_{i=1}^{\infty} \mu(A_i). The numbers Vn are non-negative, increasing in n and are all <= the infinite sum of μ(Ai). What does that tell you?

RGV
 
Yes, I think I can show that μ(A∪B)≤μ(A)+μ(B).

What is Vn, though?
 
Yagoda said:
Yes, I think I can show that μ(A∪B)≤μ(A)+μ(B).

What is Vn, though?

I *defined* Vn to be \mu \left( \cup_{i=1}^n A_i \right),; did you miss the \equiv sign?

RGV
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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