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

3 Answers 3

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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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  • $\begingroup$ @user1445572 Happy to help! $\endgroup$ Feb 3, 2013 at 18:15
  • $\begingroup$ 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? $\endgroup$
    – Prism
    Nov 4, 2013 at 2:24
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    $\begingroup$ @Prism No need for completeness. Since $A_j$ and $E$ are both measurable, $E \cap A_j$ is measurable too. $\endgroup$ Nov 4, 2013 at 9:38
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    $\begingroup$ @Ayman Hourieh: What if the function takes the value $\infty$ in the set of measure zero? We then have a product $\infty \cdot 0$ $\endgroup$
    – MSIS
    Jul 24, 2020 at 22:38
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    $\begingroup$ @MSIS The convention in this context is $\infty \cdot 0 = 0$. See Rudin's Real and Complex Analysis, page 18. $\endgroup$ Jul 25, 2020 at 10:23
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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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  • $\begingroup$ Why isn't the integral strictly less than zero? Could we use the infimum to get the other inequality? Thanks for helping! $\endgroup$ Feb 3, 2013 at 18:13
  • $\begingroup$ @user1445572: Yes, that is one way; I just added the easier way around using the infimum. $\endgroup$
    – Clayton
    Feb 3, 2013 at 18:14
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    $\begingroup$ A proof which rely on a convention is not very sastifying, is it ? $\endgroup$
    – Lierre
    Jun 9, 2013 at 9:25
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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 $

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    $\begingroup$ An additional answer is always welcome! Thanks for contributing. :D $\endgroup$ Feb 22, 2021 at 15:59
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    $\begingroup$ A beautiful answer! $\endgroup$
    – ashpool
    Jun 17, 2021 at 21:30
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    $\begingroup$ Theorem 1 is somehow equivalent to the main result in the sense that we can prove one by using the other one. The author didn't give a proof of Theorem 1 here though he promised there would be one... Let me finish his argument by giving a proof of Theorem 1 without using the main result. Proof: Suppose $\int f d\mu > \int g d\mu$. Then, $\int (f-g) d\mu > 0$. $\Longrightarrow$ There exists a set with positive measure s.t. $f-g>0$ on this set. $\Longrightarrow$ A contradiction. $\endgroup$
    – Sam Wong
    Feb 2, 2023 at 8:19
  • $\begingroup$ @SamWong Yeah that's what I notice too. I'm asking about a generalisation here 'Almost sure equality implies same expectation' is equivalent to 'integral over null set is zero.' What's equivalent to 'almost sure implies (...)'? $\endgroup$
    – BCLC
    Feb 7, 2023 at 8:34

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