Integrating f Over a Set: Is it Possible?

[JG89]

Hey guys, I have a quick question.

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

If $$E \subset S$$ then does it follow that f is also integrable over E?

 

[micromass]

Hi JG89!

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

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

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

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

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 $E=[0,1]\cap \mathbb{Q}$. 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.

 

[JG89]

Sorry micromass, I forgot one condition: that E must not be of measure zero. Are there any examples with this extra hypothesis?

 

[micromass]

I think the condition should be that E is measurable. Then it's true...

 

[JG89]

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?

 

[micromass]

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.

 

[JG89]

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!