# Integral over null set is zero

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!

• the other way around
– leo
Feb 5, 2013 at 13:52
• You do not have to be sorry. Elementariness is not an absolute concept :) Sep 7, 2015 at 11:44

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.

• @user1445572 Happy to help! Feb 3, 2013 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? Nov 4, 2013 at 2:24
• @Prism No need for completeness. Since $A_j$ and $E$ are both measurable, $E \cap A_j$ is measurable too. Nov 4, 2013 at 9:38
• Ah, of course! Thank you. :) Nov 4, 2013 at 9:44
• @MSIS The convention in this context is $\infty \cdot 0 = 0$. See Rudin's Real and Complex Analysis, page 18. Jul 25, 2020 at 10:23

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

• Why isn't the integral strictly less than zero? Could we use the infimum to get the other inequality? Thanks for helping! Feb 3, 2013 at 18:13
• @user1445572: Yes, that is one way; I just added the easier way around using the infimum. Feb 3, 2013 at 18:14
• A proof which rely on a convention is not very sastifying, is it ? Jun 9, 2013 at 9:25

Hello this was asked over 7 years ago so I am sure that this is most definitely no longer relevant for you in particular but I thought I would answer this for people in the future who might was a slightly different answer. Infact that statement holds for any measure not just Lebesgue.

Firstly a little about the notation I will use:
$$\mathbb{1}_A$$ is the "indicator function" of the set $$A$$ you may have instead seen this as $$\chi_A$$ or $$\mathcal{1}_A$$ or something else. It simply takes the value $$1$$ on $$A$$ and $$0$$ otherwise.

This proof will require:

Theorem 1:
If $$f = g$$ $$\mu$$ almost everywhere then $$\int f d\mu = \int g d\mu$$

(A proof of this theorem can be found at the bottom of this answer but this is a fairly simple result in most measure theory courses.)

Proof of main result

We wish to show that $$\int_A f d \mu = 0$$ whenever $$\mu(A) = 0$$

Firstly $$\int_A f d \mu = \int f \mathbb{1}_A d\mu$$

Secondly the function $$g := f \mathbb{1}_A$$ is $$0$$ almost everywhere as by the definition of indicator functions it is non zero only on $$A$$ which again has measure zero from the question.

More formally let us define $$N := \{ x \in X | g(x) \not = 0 \}$$ any $$x \in N$$ must be in $$A$$ because $$g(x)$$ for $$x \not \in A = 0$$ hence $$N \subseteq A$$ Hence $$\mu(N) = 0$$ (monotonicity of measure)

Hence $$g = 0$$ almost everywhere and so by Theorem 1:
$$\int_A f d\mu = \int f \mathbb{1}_A d \mu = \int g d\mu =$$(Theorem 1)$$= \int 0 d\mu = 0$$

• An additional answer is always welcome! Thanks for contributing. :D Feb 22, 2021 at 15:59
• A beautiful answer! Jun 17, 2021 at 21:30