Re: Lebesgue mersurable functions


smn wrote:
> David C. Ullrich wrote:
> > On 26 Nov 2005 23:31:47 -0800, mskirvin@xxxxxxxxx wrote:
> >
> > >nymphcc@xxxxxxxxxxx wrote:
> > >> If f(x), x in R^1, is continuous at almost every point of an
> > >> interval[a,b],
> > >> show that f is measurable on [a,b].
> > >> How to prove this statement?
> > >> Thanks a lot!
> > >
> > >So I'm assuming that f: [a,b] -> R. However, to prove measurability of
> > >the function, you need to specify the sigma algebra for the domain and
> > >range. From the heading, I'm assuming you want to prove this function
> > >is either (Lebesgue, Borel)-measurable or (Lebesgue,
> > >Lebesgue)-measurable. I think it's common to assume the sigma algebra
> > >of the range space is the Borel sigma algebra if it's not specified, so
> > >that's what I'll assume (i.e., you want to prove it's (Lebesgue,
> > >Borel)-measurable).
> > >
> > >Let E be the subset of [a,b] where f is continuous, and F the subset
> > >where it is discontinuous, so [a, b] = E \/ F. In general, if X = A \/
> > >B where A and B are measurable, then a real-valued function f is
> > >measurable on X iff it's measurable on A and on B. We will use this
> > >fact to prove that f is measurable.
> > >
> > >Since the restriction of f to E is continuous, it is measurable on E
> > >(the open sets generate the Borel measure and by defintion the inverse
> > >image of an open set under a continuous map is open). Restricting f to
> > >F, the inverse image of a measurable set is a subset of F. Since
> > >Lebesgue measure is complete and the measure of F is zero, every subset
> > >of F is also measurable with measure zero. Therefore f is also
> > >measurable on F, and hence measurable on [a, b].
> > >
> > >As a quick note, the completeness of Lebesgue measure seemed really
> > >important here, so I kind of doubt that f is necessarily (Borel,
> > >Borel)-measurable, although I'm not sure.
> >
> > That's right.
> >
> > Say K is a closed set of measure 0 and cardinality c (the Cantor
> > set, for example). Since there are only c Borel sets, K has a
> > subset E which is not a Borel set. Let f be the characteristic
> > function of E.
> >
> > Then f is not Borel measurable, but f is continuous almost
> > everywhere, since it's continuous on the complement of K.
>
> I don't think so.f as a function on all of [a,b] is required to be
> continuous at any point x of the complement of K.But such x will in
> general be limit points of K so the function is not continuous at x.
> Assume the function of the original post is bounded and continuous at
> x for a.e x in [a,b}.Then as you say in your next post ,f is Riemann
> Integrable.Using lower Riemann sums one constructs a sequence (even
> increasing) of step functions (constant on subintervals) which converge
> everywhere ,say to g and converge to f at every point of continuity of
> f .g is a borel function and I am pretty sure it is Riemann integrable
> also ,and if so g is continuous ae and g=f ae ,I still do not know
> whether f itself need be Borel measurable and I doub't it ,although the
> continuity of f at points of the complement of a set K of measure 0
> severely restricts the values of f on K .Regards,Stuart M Newberger
Whoops,sorry you did say K was closed so f is continuous on the open
complement of K.So f=char fn of E is non borel continuous ae function
=ae to the char fn of K ,a Borel function.Sorry smn

.

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