# Integral over a set of measure 0

1. Sep 20, 2010

### Boot20

Is the integral over a set of measure zero always equals to zero? Can the integral be undefined?

2. Sep 20, 2010

### mathman

It may be undefined if the function itself is peculiar with infinity as its value. For ordinary functions the integral will be 0.

Last edited: Sep 20, 2010
3. Sep 20, 2010

### wayneckm

Sorry, I have the question that, if Lebesgue integration, they always define the convention $$\infty \cdot 0 = 0$$, so, in this case, even the function takes $$\infty$$ in a set of measure 0, the integral is still 0?

4. Sep 21, 2010

### mathman

Convention is an easy way out.

5. Sep 21, 2010

### Hurkyl

Staff Emeritus
For Lebesgue integration, that the integral over a set of measure zero is a rather trivial theorem, following from the fact that all simple functions have integral zero -- so via (what I believe is) the usual formulation, it doesn't even need to be treated as a special case.

Riemann integration assumes the function is real-valued, so it doesn't even apply if you are considering extended-real-number-valued functions that take on the values $+\infty$ or $-\infty$.

Last edited: Sep 21, 2010
6. Sep 21, 2010

### wayneckm

But apparently to me $$\infty \cdot 0 = 0$$ should be adopted
Else, if $$f$$ admit $$\infty$$ on set $$A$$ of measure $$0$$, we may use $$f_{n} = n$$ on $$A$$ to approximate $$f$$ from below, then, the integral of $$f_{n}$$ is zero, by monotone convergence theorem, the integral of $$f$$ should be zero as well. If we do not define $$\infty \cdot 0 = 0$$, we may get inconsistency in this case?

7. Sep 22, 2010

### Boot20

But what if the f_{n} are defined over sets of measure non-zero, but that the sum of the measure of those sets converges to zero?

Last edited: Sep 22, 2010
8. Sep 22, 2010

### wayneckm

Sorry that I cannot get your idea exactly. Or can you tell me explicitly what is the function $$f$$ to be integrated?