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From: agapito6314 on 14 Jun 2010 15:16
For continuous f: R ---> R, how does one prove measurability using
only 2 basic concepts:

Continuity of f at c: For each e, there exists d such that

| x - c | < d -----> | f(x) - f(c) | < e

Measurability of f : For all real a

{ x: f(x) > a} is open (or Borel) set.

Many thanks for all help.


From: Arturo Magidin on 14 Jun 2010 15:26
On Jun 14, 2:16 pm, agapito6...(a)aol.com wrote:
> For continuous  f: R ---> R, how does one prove measurability using
> only 2 basic concepts:
>
> Continuity of f at c:  For each e, there exists d such that
>
> | x - c | < d   ----->  | f(x) - f(c) | < e
>
> Measurability of f :  For all real a
>
> { x: f(x) > a} is open (or Borel) set.
>
> Many thanks for all help.

Let c be an element of S_a = {x : f(x) > a}.

Let e = (f(c)-a)/2; then e>0 since c is in S_a.

Thus, there exists d>0 such that for all x in (c-d,c+d), |f(x)-f(c)|
<e; in particular, f(c-d,c+d) is contained in (f(c)-e,f(c)+e), which
in turn is contained in (a,oo), since f(c)-e > a.

Therefore, (c-d,c+d) is contaied in S_a.

Thus, for every c in S_a, there is an open interval containing c and
completely contains in S_a. Thus, S_a is open.

Hence, f is measurable.

--
Arturo Magidin

From: agapito6314 on 14 Jun 2010 16:06
On Jun 14, 2:26 pm, Arturo Magidin <magi...(a)member.ams.org> wrote:
> On Jun 14, 2:16 pm, agapito6...(a)aol.com wrote:
>
> > For continuous  f: R ---> R, how does one prove measurability using
> > only 2 basic concepts:
>
> > Continuity of f at c:  For each e, there exists d such that
>
> > | x - c | < d   ----->  | f(x) - f(c) | < e
>
> > Measurability of f :  For all real a
>
> > { x: f(x) > a} is open (or Borel) set.
>
> > Many thanks for all help.
>
> Let c be an element of S_a = {x : f(x) > a}.
>
> Let e = (f(c)-a)/2; then e>0 since c is in S_a.
>
> Thus, there exists d>0 such that for all x in (c-d,c+d), |f(x)-f(c)|
> <e; in particular, f(c-d,c+d) is contained in (f(c)-e,f(c)+e), which
> in turn is contained in (a,oo), since f(c)-e > a.
>
> Therefore, (c-d,c+d) is contaied in S_a.
>
> Thus, for every c in S_a, there is an open interval containing c and
> completely contains in S_a. Thus, S_a is open.
>
> Hence, f is measurable.
>
> --
> Arturo Magidin

As always, many thanks Prof. Magidin.
From: adamk on 14 Jun 2010 12:08
> For continuous f: R ---> R, how does one prove
> measurability using
> only 2 basic concepts:
>
> Continuity of f at c: For each e, there exists d
> such that
>
> | x - c | < d -----> | f(x) - f(c) | < e
>
> Measurability of f : For all real a
>
> { x: f(x) > a} is open (or Borel) set.
>
> Many thanks for all help.
>
>

why not just use the fact that continuity is stronger
than measurability: let U be open in R, then U is Borel-measurable. Since f is continuous, f^-1(U) is open in R,
and so f^-1(U) is measurable, so f itself is measurable.
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