A function that is Lebesgue integrable but not measurable (not absurd obviously)

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.