'Almost everywhere' proof

jinsing · Nov 18, 2011

1. The problem statement, all variables and given/known data

Prove the following: If $\{f_n\}$ is a collection of measurable functions defined on $\rr$ and satisfying:

1) $f_n(x)\le 1 \, \forall n \in \nn \, \forall x \in \rr$ and
2) $f_n(x) \ge 0 \mbox{ a.e. on } \rr \, \forall n \in \nn$ and
3) $\ds f(x)=\sup\{f_n(x) \mid n \in \nn\}$,

then $f(x) \ge 0$ a.e. on $\rr$.

2. Relevant equations

Definition of 'almost everywhere,' knowledge of measurability, supremums etc.

3. The attempt at a solution

I've been racking my brain trying to figure out..what's going on in this problem - I'm not too good at comprehending 'almost everywhere.' I think I want to show that \{ x \in \rr | f(x) > 0\} has measure 0, but I don't really know how to get there. I began by substituting, so I'd be showing \{x \in \rr | \sup\{f_n(x) | n \in \nn\} > 0\} has measure 0, but that seems too messy and convoluted to be correct.

I think this is a relatively straight-forward proof, but I'm not entirely sure how to get started. A gentle push in the right direction would be great.

Dick · Nov 18, 2011

I really can't read that. I can sort of guess what some of it means, but I'm not sure I get the whole point. The answer might be that the countable union of measure zero sets has measure zero. But that's just a guess. Can you either write that out in words or TeX it up properly. See https://www.physicsforums.com/showthread.php?t=8997 The $ $ thing doesn't work here. You have to use the [ tex ] [ / tex ] thing (without the spaces).

jinsing · Nov 18, 2011

Oh okay sorry. Here it is:

Prove the following: If [tex]\{f_n\} [/tex] is a collection of measurable functions defined on [tex]R[/tex] and satisfying:

1) [tex]f_n(x)\le 1 \, \forall n \in N \, \forall x \in R[/tex] and
2) [tex]f_n(x) \ge 0 \mbox{ a.e. on } R \, \forall n \in N[/tex] and
3) [tex] f(x)=\sup\{f_n(x) \mid n \in N\}[/tex],

then [tex]f(x) \ge 0[/tex] a.e. on [tex]R[/tex].

Thanks for your time.

Dick · Nov 18, 2011

No need to apologize. That helps. Pick S_n to be the set of all x such that f_n(x)<0. That has measure zero, right? Define S to be the union of all of the S_n. What can you say about f(x) outside of S?

jinsing · Nov 19, 2011

Then can we conclude that f(x) >= 0 for all x \in R-S, and since S has measure zero, f holds this property almost everywhere? Or is there something else I'm missing?

Dick · Nov 19, 2011

jinsing said: ↑

Then can we conclude that f(x) >= 0 for all x \in R-S, and since S has measure zero, f holds this property almost everywhere? Or is there something else I'm missing?

That's pretty much it, in outline.

jinsing · Nov 19, 2011

Oh, well thanks then! Didn't think it'd be that straightforward.

Bacle2 · Nov 19, 2011

One point to consider here is why we can talk about f being a.e≥ 0 , i.e., why we can assign a measure to f^-1[0,∞) . This is because the sup of a collection of measurable function is itself measurable. Sorry, I'm running, I'll come back .

Dick · Nov 19, 2011

Bacle2 said: ↑

One point to consider here is why we can talk about f being a.e≥ 0 , i.e., why we can assign a measure to f^-1[0,∞) . This is because the sup of a collection of measurable function is itself measurable. Sorry, I'm running, I'll come back .

I'm not sure I get your concern. All measure zero sets are measurable. The complement of a measurable set is measurable. So? I don't even think you need to assume f_n is measurable. What do you think, jinsing?

jinsing · Nov 21, 2011

Yeah, I suppose if I were really crossing every 't' and dotting every 'i' I could mention that, but I think the implications of measurability are straightforward enough.

Bacle2 · Nov 21, 2011

I hope I'm not being too much of an egghead here, but actually, what I was thinking was the issue of subsets of sets of measure zero, which are not necessarily measurable, unless you are working with a complete measure; if you describe the set where f>0 as S={x:f_n(x)>0} for some natural n, then you do run into subsets of sets of measure zero; but if f is measurable, this is not a problem, (because of the description of S), whether the measure is complete or not.

Dick · Nov 22, 2011

Bacle2 said: ↑

I hope I'm not being too much of an egghead here, but actually, what I was thinking was the issue of subsets of sets of measure zero, which are not necessarily measurable, unless you are working with a complete measure; if you describe the set where f>0 as S={x:f_n(x)>0} for some natural n, then you do run into subsets of sets of measure zero; but if f is measurable, this is not a problem, (because of the description of S), whether the measure is complete or not.

Yeah, but i) I kind of thought we were dealing with Lesbegue measure on R, and that is complete. And ii) I think the countable union of measure zero sets is still measure zero, regardless of whether the measure is complete or not. And (iii) who claimed a subset of a measure zero set has measure zero anyway?

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'Almost everywhere' proof

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