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 Posted: Wed, 15 Feb 2012 19:05:33 UTC
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Hi, my intuition tells me that should be a continuous function of x if is continuous almost everywhere. I'm not asking that someone should prove this for me (if possible), but I'm wondering whether anyone could point me towards a reference where this is shown. I have been searching and not finding anything. Thanks!

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 Posted: Wed, 15 Feb 2012 19:15:52 UTC
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nicster2009 wrote:
Hi, my intuition tells me that should be a continuous function of x if is continuous almost everywhere. I'm not asking that someone should prove this for me (if possible), but I'm wondering whether anyone could point me towards a reference where this is shown. I have been searching and not finding anything. Thanks!

No, let

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 Posted: Wed, 15 Feb 2012 23:58:53 UTC
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Thanks. But what if f is bounded? If it makes a difference, in my case the domains of x and y are closed intervals and f is real valued.

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 Posted: Thu, 16 Feb 2012 04:11:36 UTC
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nicster2009 wrote:
Thanks. But what if f is bounded? If it makes a difference, in my case the domains of x and y are closed intervals and f is real valued.

Assuming finiteness of the measure you put on the closed interval (on which y is allowed to reside), it is just a matter of using uniform continuity of f.

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 Posted: Thu, 16 Feb 2012 12:20:16 UTC
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But I don't even assume that f is continuous...

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 Posted: Thu, 16 Feb 2012 21:24:23 UTC
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nicster2009 wrote:
But I don't even assume that f is continuous...

You have that is in for , then since is ctns a.e., it's equivalence class in contains a continuous function, so you may as well assume is continuous for the purpose of integration.

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 Posted: Fri, 17 Feb 2012 03:25:31 UTC
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nicster2009 wrote:
But I don't even assume that f is continuous...

The answer is no. For example, let for some . Then a.e. continuity of f is equivalent to is null. But then is not continuous unless A is ...

On the other hand, you have is continuous almost everywhere. (Try to prove it!)

nicster2009 wrote:
But I don't even assume that f is continuous...

You have that is in for , then since is ctns a.e., it's equivalence class in contains a continuous function, so you may as well assume is continuous for the purpose of integration.

Actually no. For example, a step function is continuous almost everywhere (since there is only discontinuous at the end-point of the interval), but admits no continuous representative.

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 Posted: Fri, 17 Feb 2012 04:49:43 UTC
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outermeasure wrote:
nicster2009 wrote:
But I don't even assume that f is continuous...

The answer is no. For example, let for some . Then a.e. continuity of f is equivalent to is null. But then is not continuous unless A is ...

On the other hand, you have is continuous almost everywhere. (Try to prove it!)

nicster2009 wrote:
But I don't even assume that f is continuous...

You have that is in for , then since is ctns a.e., it's equivalence class in contains a continuous function, so you may as well assume is continuous for the purpose of integration.

Actually no. For example, a step function is continuous almost everywhere (since there is only discontinuous at the end-point of the interval), but admits no continuous representative.

You're right, and I see exactly where I was off in my thinking, thanks for that outermeasure!

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 Posted: Fri, 17 Feb 2012 19:14:39 UTC
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Joined: Wed, 15 Feb 2012 18:56:18 UTC
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Thanks a lot. I had been working on a more restricted version of the same problem a while ago and from that I took away that it should hold. The counter example is quite straightforward though. I should have thought a bit more about the general case before posting.

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