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I think: A function $f$, as long as it is measurable, though Lebesgue integrable or not, always has Lebesgue integral on any domain $E$.

However Royden & Fitzpatrick’s book "Real Analysis" (4th ed) seems to say implicitly that “a function could be integrable without being Lebesgue measurable”. In particular, theorem 7 page 103 says:

“If function $f$ is bounded on set $E$ of finite measure, then $f$ is Lebesgue integrable over $E$ if and only if $f$ is measurable”.

The book spends a half page to prove the direction “$f$ is integrable implies $f$ is measurable”! Even the book “Real Analysis: Measure Theory, Integration, And Hilbert Spaces” of Elias M. Stein & Rami Shakarchi does the same job!

This makes me think there is possibly a function that is not bounded, not measurable but Lebesgue integrable on a set of infinite measure?

=== Update: Read the answer of smnoren and me below about the motivation behind the approaches to define Lebesgue integrals. Final conclusion: The starting statement above is still true and doesn't contradict with the approach of Royden and Stein.

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    $\begingroup$ The function $f(x)=1/x$ for $x\ne0$, $f(0)=0$ is measurable, but it has no integral over $\mathbb{R}$ (even infinite). $\endgroup$
    – egreg
    Jun 23, 2015 at 23:14
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    $\begingroup$ @egreg I think the OP seeks to understand why the text considers the possibility of the other direction: integrable but not measurable $\endgroup$
    – angryavian
    Jun 23, 2015 at 23:18
  • $\begingroup$ @egreg: you understand wrong my question. I need a function that is integrable but not measurable. $\endgroup$
    – Thang
    Jun 23, 2015 at 23:18
  • $\begingroup$ @Thang My comment was about your starting statement. $\endgroup$
    – egreg
    Jun 23, 2015 at 23:19

3 Answers 3

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There is a subtle difference in defining Lebesgue integrals in Real analysis textbooks:

I) The approach of Royden & Fitzpatrick (in “Real analysis” 4th ed), Stein & Shakarchi (in “Real Analysis: Measure Theory, Integration, And Hilbert Spaces”)

Firstly, it defines Lebesgue integrability and Lebesgue integral for a bounded function (not necessarily measurable) on a domain of finite measure. A bounded function needs to be Lebesgue integrable first (the upper and the lower Lebesgue integral agree), then the integral can be defined to be this common value. The authors’ motivation is try to define “Lebesgue integrability” like “Rieman integrability”: upper integral equals lower integral.

However, unfortunately, the upper and lower Lebesgue integrals don’t agree for an arbitrary Lebesgue integrable function, so when the authors move to functions in general (not necessarily bounded), they still have to go back to the requirement "measurable". This sudden appearance of "measurability" is not natural.

(Note that the upper/lower Darboux sum in the definition of Rieman integrability can be viewed as step functions, which are a special case of simple functions. So “upper/lower Rieman (Darboux) integral” is a special case of “upper/lower Lebesgue integral”)

II) The approach of Folland (in “Real Analysis: Modern Techniques and Their Applications”), Bruckner & Thomsom (in “Real analysis”), Carothers (in “Real analysis”), etc.

The construction requires a function to be measurable, and defines the Lebesgue integral to be the upper Lebesgue integral, and when the integral is finite the function is said to be Lebesgue integrable.

This approach doesn’t immediately show how Lebesgue integral convers Rieman integral, so later on, the author proves that in the case a function is bounded and the domain of integration is of finite measure: the upper Lebesgue integral equals to the lower Lebesgue integral, which means Lebesgue integral is reduced to Rieman integral.

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On pg. 73 of Royden & Fitzpatrick, Lebesgue integrability is defined for bounded functions on domains of finite measure, without the assumption of measurability. However, this theorem that you have reveals that functions of this type can't be integrable unless they are measurable. Hence, the definition of integrable on pg. 73 is consistent with all the other definitions in the book that assume the function to be measurable.

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  • $\begingroup$ This is the correct answer: if you make additional hypotheses about the functions, you can define the Lebesgue integral without needing to be able to measure preimages. It just turns out that if the upper and lower integral agree then you necessarily can measure preimages. $\endgroup$
    – Ian
    Jun 24, 2015 at 1:00
  • $\begingroup$ Unfortunately, the upper Lebesgue and lower Lebesgue integral don't agree when $f$ is not bounded on the domain of integration, which is probably not of finite measure. They agree only in the case $f$ is bounded and the domain is of finite measure. $\endgroup$
    – Thang
    Jun 24, 2015 at 1:08
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Let $A\subset [0,1]$ be a nonmeasurable set. Then consider the function $f(x)=1$, if $x\in A$ and $f(x)=-1$, if $x \in [0,1]\setminus A$ and zero everywhere else. Then clearly $f$ is not measurable, but $|f(x)|=\chi_{[0,1]}(x)$, so it's integrable.

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    $\begingroup$ Since your function is bounded, so by the theorem above, which is very well known, the function must be measurable - which is contradictory. If $|f|$ is integrable, then $f$ is not necessarily integrable. See example at math.stackexchange.com/questions/1334374/… $\endgroup$
    – Thang
    Jun 23, 2015 at 23:49
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    $\begingroup$ @TyrCurtis The definition of Lebesgue integrable that you are using requires the function $f$ to be measurable, which it clearly isn't. See the definition on pg. 85. $\endgroup$
    – srnoren
    Jun 24, 2015 at 0:32
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    $\begingroup$ @Thang I agree with your second statement that $|f|$ can be integrable but $f$ is not (which is his case). However, this is example is not contradictory and the function is not measurable why? because in the theorem assumption, it requires $E$ to be a measurable set of finite measure and the domain of $f$ is $A\cup [0,1]\setminus A$ which is not a measurable set. $\endgroup$ Sep 17, 2017 at 22:31
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    $\begingroup$ The last sentence in this answer is ambiguous: "so it's integrable" could refer to $|f|$ or it could refer to $f$. The correct unambiguous statement is that $f$ is not integrable and $|f|$ is integrable. $\endgroup$
    – jdods
    May 7, 2021 at 18:35

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