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I am sorry for this elementary question, but i could not figure out a rigorous proof for why the Lebesgue integral of any function over a null set is zero.

Thanks for helping!

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the other way around –  leo Feb 5 '13 at 13:52

2 Answers 2

up vote 7 down vote accepted

Consider that $$\int_E f(x)\leq \sup|f(x)|\cdot m(E).$$ With the convention that $\infty\cdot 0=0$, we have $$\left|\int_E f(x)\right|\leq \sup|f(x)|\cdot m(E)=0.$$

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Why isn't the integral strictly less than zero? Could we use the infimum to get the other inequality? Thanks for helping! –  Guilherme Salomé Feb 3 '13 at 18:13
    
@user1445572: Yes, that is one way; I just added the easier way around using the infimum. –  Clayton Feb 3 '13 at 18:14
    
Awesome, thanks a lot! =] –  Guilherme Salomé Feb 3 '13 at 18:15
4  
A proof which rely on a convention is not very sastifying, is it ? –  Lierre Jun 9 '13 at 9:25

Start with the definition. The Lebesgue integral of a simple function $s = \sum_{j=1}^n \alpha_j \ \chi_{A_j}$ is:

$$ \int_E s \,d\mu = \sum_{j=1}^n \alpha_i \ \mu(E \cap A_j) $$

If $\mu(E) = 0$, then $\mu(E \cap A_j) = 0$ for all $j$. Thus $\int_E s \,d\mu = 0$.

The Lebesgue integral of a nonnegative function $f$ is the supremum of integrals of all simple functions $s$ such that $0 \le s \le f$. Since all of these integrals are $0$, the supremum is $0$ too.

Since every real function $f$ can be written as $f = f^+ - f^-$ where $f^+$ and $f^-$ are both nonnegative, we have $\int_E f \, d\mu = 0$ too. The general result follows from the fact that every complex function $f$ can be written as $f = u + i v$ where $u$ and $v$ are real.

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Thanks Ayman! Got it! –  Guilherme Salomé Feb 3 '13 at 18:14
    
@user1445572 Happy to help! –  Ayman Hourieh Feb 3 '13 at 18:15
    
When going from $\mu(E)=0$ to $\mu(E\cap A_{j})=0$, aren't you assuming that the measure space is complete? If so, is there any way to modify the proof if $X$ is not complete? –  Prism Nov 4 '13 at 2:24
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@Prism No need for completeness. Since $A_j$ and $E$ are both measurable, $E \cap A_j$ is measurable too. –  Ayman Hourieh Nov 4 '13 at 9:38
    
Ah, of course! Thank you. :) –  Prism Nov 4 '13 at 9:44

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