Integrating f Over a Set: Is it Possible?

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Discussion Overview

The discussion revolves around the integrability of a function f over a subset E of a set S, where S is integrable. Participants explore conditions under which f remains integrable over E, focusing on concepts related to Riemann and Lebesgue integration.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether a function f that is integrable over a set S implies that it is also integrable over any subset E of S.
  • Another participant argues that this is not necessarily true, providing an example where f is integrable over [0,1] but not over the set of rational numbers within that interval.
  • A later reply introduces the condition that E must not be of measure zero and asks for examples under this hypothesis.
  • Another participant suggests that E should be measurable for the integrability of f to hold true.
  • There is a clarification that measurable sets can include those of measure zero, but this does not guarantee Riemann integrability.
  • One participant expresses disappointment that the integrability condition does not hold even when excluding measure zero sets, citing an example involving the union of rationals and an interval.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the conditions under which f remains integrable over E. Multiple competing views regarding the definitions and implications of integrability persist throughout the discussion.

Contextual Notes

The discussion highlights the complexity of integrability, particularly distinguishing between Riemann and Lebesgue integration, and the implications of set measurability and measure zero on these concepts.

JG89
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Hey guys, I have a quick question.

Suppose we have a function f that is integrable over a set [tex]S \subset \mathbb{R}^n[/tex]

If [tex]E \subset S[/tex] then does it follow that f is also integrable over E?
 
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Hi JG89! :smile:

This is sadly enough not true. I'll try to explain it without going in too much detail.

The point is that the set E can be very, very ugly. Consider, for example, the function

[tex]f:[0,1]\rightarrow \mathbb{R}:x\rightarrow 1[/tex]

this is a very innocent function and is certainly integrable over [0,1]. However, if I take

[tex]E=[0,1]\cap \mathbb{Q}[/tex]

then f is not integrable over E anymore (Riemann-integrable that is). The reason is that E is far too ugly.

One can solve this issue by allowing more sets E, and this yields the Lebesgue integral. This resolves the issue with [itex]E=[0,1]\cap \mathbb{Q}[/itex]. Sadly, the issue cannot be entirely resolved, as there will be (extremely ugly) sets E over which f cannot be integrable. Luckily enough, these ugly sets won't occur in daily practise. For example, if E is open or closed or the union of open/closed sets, then f will remain integrable over E. But it's important to know that there exists ugly sets over which f cannot be integrated.
 
Sorry micromass, I forgot one condition: that E must not be of measure zero. Are there any examples with this extra hypothesis?
 
I think the condition should be that E is measurable. Then it's true...
 
By measurable you just mean that the set can be assigned a measure, right? So even a set of measure zero would be measurable then, right? So then what do you mean that if the set is measurable then it is true?
 
JG89 said:
By measurable you just mean that the set can be assigned a measure, right? So even a set of measure zero would be measurable then, right? So then what do you mean that if the set is measurable then it is true?

Yes, for Lebesgue integration, if the set can be assigned a measure (even measure zero), then what you say is true.
For Riemann integration, however, it is not true. Not even if you exclude sets of measure zero. For example, take [0,2] and take [itex]E=\mathbb{Q}\cup [0,1][/tex], then f is not0 Riemann-integrable over E and E does not have measure zero.[/itex]
 
Damn, I was getting my hopes up that maybe it would be true if E weren't of measure zero! Thanks for the help though!
 

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