# 'Almost everywhere' proof

## Homework Statement

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 $\rr$.

## Homework Equations

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

## 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
Homework Helper
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).

Oh okay sorry. Here it is:

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

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

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

Dick
Homework Helper
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?

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
Homework Helper
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.

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

Bacle2
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
Homework Helper
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?

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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