Borel sets of the extended real line

[sto]

Define a real valued function f on a measurable space (X,S) to be
*measurable* if, for any Borel set M,

intersection({x:f(x) != 0}, {x:f(x) in M}) in S

It is my understanding that this definition can be generalized to an
extended real valued function g on a measurable space (X,S) by
regarding the two subsets {+oo} and {-oo} of the extended real line R_
to be Borel sets.

1. Does that mean that unions and differences of sets such as, say,
{-oo} union M, or such as {-oo} \ M, where M is a Borel set of the
real line R, are also Borel sets? In other words, are the Borel sets
of the extended real line, including {+oo} and {-oo}, still a ring?

2. Is it the case that the extended real line

R_ = {-oo} union R union{+oo}

is itself a Borel set of the extended real line and therefore, given a
measurable extended real valued function g on (X,S), that

intersection({x:g(x) != 0},{x, g(x) in R_}) =
intersection({x:g(x) != 0}, X) =
{x:g(x) != 0} in S

in which case {x:g(x) != 0} is a measurable set?

3. Going further, if h is an extended real valued Borel measurable
function on the extended real line, then is it the case that for any
Borel set M of the extended real line

intersection({t:h(t) != 0},{t:h(t) in M})

will be a Borel set of the extended real line (as opposed to a Borel
set of the regular real line, which set does not contain the elements
+oo or -oo)?

Thanks,
-sto

[Dave L. Renfro]

sto <s...@address.invalid> wrote (in part):

> It is my understanding that this definition can be generalized
> to an extended real valued function g on a measurable space
> (X,S) by regarding the two subsets {+oo} and {-oo} of the
> extended real line R_ to be Borel sets.

It seems to me that your questions can be simply answered by
observing {+oo} and {-oo} are closed sets (of a very simple
type, in fact, being singletons), which are about as simple
as you can be for a Borel set. Note also that the real line
is an open subset of the extended real line. In fact, without
looking back to be sure, I think all the sets you mentioned
are not just Borel, but in fact belong to the lowest non-trivial
ambiguous class in the Borel hierarchy (kind of like saying
a number is less than a googleplex when it's actually either
0 or 1, except it's even more extreme than this).

Dave L. Renfro

[Dave L. Renfro]

Dave L. Renfro wrote (in part):

> In fact, without looking back to be sure, I think all
> the sets you mentioned are not just Borel, but in fact
> belong to the lowest non-trivial ambiguous class in the

> Borel hierarchy [...]

O-K, I re-read your post and I see mention of measurable
functions and sets obtained from them, so in these cases
you could be arbitrarily high in the Borel hierarchy, or
even beyond it (unless by "measurable" you mean "Borel
measurable"). My earlier comment was based on some statements
you made, such as the following:

"Is it the case that the extended real line

R_ = {-oo} union R union{+oo}

is itself a Borel set of the extended real line"

Any topological space is an open subset of itself,
as well as a closed subset of itself.

Dave L. Renfro

[sto]

Dave L. Renfro wrote:
> sto <s...@address.invalid> wrote (in part):
>
>> It is my understanding that this definition can be generalized
>> to an extended real valued function g on a measurable space
>> (X,S) by regarding the two subsets {+oo} and {-oo} of the
>> extended real line R_ to be Borel sets.
>
> It seems to me that your questions can be simply answered by
> observing {+oo} and {-oo} are closed sets (of a very simple
> type, in fact, being singletons), which are about as simple
> as you can be for a Borel set. Note also that the real line
> is an open subset of the extended real line.

Perhaps the problem is how the term "Borel set" is defined for subsets
of the extended real line. From what I've been told, to get the Borel
sets of the extended real line one takes the Borel sets of the real line
and simply tacks on two elements: +oo and -oo. What I don't understand
is how one can take a sigma-ring and simply add two sets to it and still
have a sigma ring. For example, the class {0,X} of subsets of a set X
is a sigma-ring. If I then decide to add two arbitrary sets, a and b,
to it I get the class of sets {0,a,b,X}, which is no longer necessarily
a sigma-ring (I think the set difference a\b, for example, is not member
of the class if a and b are distinct and nonempty).

If {+oo} and {-oo} are declared to be Borel sets, then how do I know
that the class of Borel sets is still a sigma-ring? In fact, how do I
know whether any other statements about the Borel sets of the real line
are true for the Borel sets of the extended real line? (for example, it
does not seem to be the case that the Borel sets of the extended real
line are the sigma ring generated by the semiring of sets of the form
{x: -oo <= c < d < oo}, for c and d real).

Thanks
-sto

[David C. Ullrich]

On Fri, 20 Mar 2009 18:07:48 -0400, sto <s...@address.invalid> wrote:

>Dave L. Renfro wrote:
>> sto <s...@address.invalid> wrote (in part):
>>
>>> It is my understanding that this definition can be generalized
>>> to an extended real valued function g on a measurable space
>>> (X,S) by regarding the two subsets {+oo} and {-oo} of the
>>> extended real line R_ to be Borel sets.
>>
>> It seems to me that your questions can be simply answered by
>> observing {+oo} and {-oo} are closed sets (of a very simple
>> type, in fact, being singletons), which are about as simple
>> as you can be for a Borel set. Note also that the real line
>> is an open subset of the extended real line.
>
>Perhaps the problem is how the term "Borel set" is defined for subsets
>of the extended real line. From what I've been told, to get the Borel
>sets of the extended real line one takes the Borel sets of the real line
>and simply tacks on two elements: +oo and -oo. What I don't understand
>is how one can take a sigma-ring and simply add two sets to it and still
>have a sigma ring.

That's not what was meant when someone said "tack on" those
two elements. The truth is this:

A is a Borel subset of [-oo,oo] if and only if there exists
B, a Borel subset of R, such that A is equal to one of
the four sets B, B union {oo}, B union {-oo} and
B union {oo, -oo}.

>For example, the class {0,X} of subsets of a set X
>is a sigma-ring. If I then decide to add two arbitrary sets, a and b,
>to it I get the class of sets {0,a,b,X}, which is no longer necessarily
>a sigma-ring (I think the set difference a\b, for example, is not member
>of the class if a and b are distinct and nonempty).
>
>If {+oo} and {-oo} are declared to be Borel sets, then how do I know
>that the class of Borel sets is still a sigma-ring? In fact, how do I
>know whether any other statements about the Borel sets of the real line
>are true for the Borel sets of the extended real line?

This should be clear, with the revised notion of "tack on oo and -oo".

>(for example, it
>does not seem to be the case that the Borel sets of the extended real
>line are the sigma ring generated by the semiring of sets of the form
>{x: -oo <= c < d < oo}, for c and d real).

The notation is scrambled - you meant "sets of the form
{x : c < x < d} for -oo < c < d < oo. (Leaving out one
"x" was a minor typo, but it's _important_ that the
restrictions on c and d are _not_ included inside those
braces...) Anway:

That's not what's required. What's required is that the Borel
algebra on [-oo, oo] should be generated by the _open_
_subsets_ of [-oo, oo].

That turns out to be the algebra generated by the open
intervals, _except_ you need to note that now
(a, oo], [-oo, b) and [-oo, oo] count as "open
intervals". (Well, they're intervals, and they're open...)

>Thanks
>-sto

David C. Ullrich

"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)

[zzzhhh]

hi, see here:
http://mathoverflow.net/questions/23826/what-is-the-associated-borel-set-of-a-borel-measurable-function-on-the-extended-r