# Integral over null set is zero - measure theory

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

Thanks for helping!

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

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 at 18:13 @user1445572: Yes, that is one way; I just added the easier way around using the infimum. – Clayton Feb 3 at 18:14 Awesome, thanks a lot! =] – Guilherme Salomé Feb 3 at 18:15

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 at 18:14 @user1445572 Happy to help! – Ayman Hourieh Feb 3 at 18:15