Convergence of a sequence of sets

In summary, the Lebesgue's Dominated Convergence Theorem says that if given a sequence of sets, each of which has an outer measure, then the limit of the sequence as n approaches infinity is the set with the largest outer measure.
  • #36
No. My problem was exactly why ##\|f(\chi(A_n))(x)-f(\chi(A))(x)\|<\varepsilon## without further conditions on ##f##.
 
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  • #37
Well we know that f is integrable and ##A_n## is measurable, is that not enough?
 
  • #38
benorin said:
Well we know that f is integrable and ##A_n## is measurable, is that not enough?
I am not sure. I suppose we need ##f## to be continuous, but maybe Lebesgue integrable will do. I just don't see it. We have that ##\|\chi(A_n)(x)-\chi(A)(x)\|## is small, but why does applying ##f## keep this condition?
 
  • #39
I found a remark in Baby Rudin that I think may help,

Remark: If f is Lebesgue integrable on E, and if ##A\in\mathfrak{M}## (A is measurable) and ##A\subset E##, then f is Lebesgue integrable on A.

Also that if f is Lebesgue integrable on E, the f is finite on E a.e. may also help, I think
 
  • #40
What if I let ##B_n:=\bigcap_{k=1}^n A_k## and then define ##f_k:=f\circ\chi (B_k)## and proceed with the proof as before, would

$$
\lim_{n \to \infty} f\circ \chi(B_n) {=}_{a.e.} f \circ \chi\left(\displaystyle{\bigcap_{k=1}^{\infty}}A_k\right)
$$

still give you pause?
 
  • #41
Sure, but what makes you confident, that ##f## doesn't destroy that behavior? If ##f## is continuous, o.k., but is it?
 
  • #42
Let ##B_n:=\bigcap_{k=1}^n A_k## and then define ##f_k:=f\circ\chi (B_k)## and proceed with the proof as before, let's examine whether

$$
\lim_{n \to \infty} f\circ \chi(B_n) {=}_{a.e.} f \circ \chi\left(\displaystyle{\bigcap_{k=1}^{\infty}}A_k\right)
$$

is true: is the measure of the set where they're not equal equal to zero? I.e. is

$$\mu \left( S\left( \lim_{n\to\infty}B_n , \bigcap_{k=1}^{\infty}A_k\right) \right)$$
$$=\mu \left\{\left( \left( \lim_{n\to\infty}B_n\right) - \bigcap_{k=1}^{\infty}A_k\right) \cup \left(\left(\bigcap_{k=1}^{\infty}A_k\right) - \left( \lim_{n\to\infty}B_n\right)\right) \right\}=0?$$

where ##S(A,B)## is the symmetric difference of sets; I think this follows from the definnition of ##B_n## trivially. So, satisfied that the ##\chi 's## are equal everywhere it follows that ##f\circ \chi 's## are also equal since applying f just inserts the value of f(x) if x is in the set ##\chi## is of (and zero otherwise) and these are equal sets on both sides so evaluating f in the same places on both sides should be equal everywhere, right?
 
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  • #43
"Let S,Sn be as in the problem statement. It is clear that S is the (increasing) union of the Sn. This is already enough to guarantee that m(S(S,Sn))=m(S−Sn) goes to zero as n→∞."

To be airtight, it's necessary to account for the boundary ∂S of S.
 
  • #44
Why? I gave an argument- do you disagree with any part of it?
 
  • #45
@zinq , I'm uncertain which post you are quoting but I should make you aware there has been at least 3 different definitions of convergence of a sequence of sets employed at various times in this thread. I ended up using point-wise convergence and the proof of post #31. As for showing that the particular sequence of sets I mentioned I needed to work with the theorem in post #31 I have used bivariate induction and three different Lagrange Multipliers proofs to flesh out the induction proof (not shown here). If you mean to point out a shortcoming of the proof in post #31, then I am most interested in what you have to say.
 

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