I am studying Measure Theory, and am a bit confused about the way that the Lebesgue Integral is defined.

Wikipedia states that:

We denote the integral of an extended Borel function $f$ as following:

$$ \int _{\Omega} f d\mu = \int _{\Omega} f^+ d\mu - \int _{\Omega}f^- d\mu $$

This is motivated by the “standard machine” where we build up the notion of the integral from indicator functions. But here there is seemingly no constraint on what our choice of Measure $\mu$ is.

Whereas, I have previously been led to believe that the Lebesgue Integral specifically refers to integration with respect to the Lebesgue Measure $m$ on $\mathbb{R}$.

I would be interested in knowing which of these these is correct or if I am possibly misunderstanding something. And if Lebesgue integration is the more general act of integration with respect to any measure, then does this not mean that we have multiple different Lebesgue integrals depending on our choice of measure?

  • $\begingroup$ Your suspicion is correct, Lebesgue measure is not a single canonical thing, it is a way to define integration on arbitrary measurable space. If you pick Lebesgue measure to be your space then you get the standard Lebesgue integral. But if you copy the definition of standard Lebesgue integral to any other measurable space, you got yourself what you would call the Lebesgue integral of that space. So yeah, Lebesgue integral is really an integration method rather than an integral intself. $\endgroup$
    – donaastor
    Dec 18, 2022 at 22:41
  • $\begingroup$ Generally the Lebesgue integral denotes the general integral with respect to some measure, however there are authors who do define the Lebesgue integral as specifically integration with respect to the Lebesgue measure (see, for example, Measure Theory by Donald L. Cohn). At the end of the day, it's just a question of convention of the terminology. $\endgroup$
    – Lorago
    Dec 18, 2022 at 22:44
  • $\begingroup$ Yes, sorry, I was much less clear then I wanted to be... Yes, you can pick your own measure. That's what I meant by arbitrary measurable space. So in your terms, that space you just wrote can really be just anything. It doesn't have to relate to real numbers in any way, nor to the Lebwegue measure in any way. $\endgroup$
    – donaastor
    Dec 18, 2022 at 22:48

1 Answer 1


We define the (Lebesgue) integral with respect to an arbitrary measure space $(X,\mathcal{M},\mu)$, where $X$ is a set, $\mathcal{M}$ is a $\sigma$-algebra of subsets of $X$, and $\mu$ is an arbitrary measure. If we let $(X,\mathcal{M},\mu)=(\mathbb{R}^n,\mathcal{L},m)$, where $m$ is the Lebesgue measure and $\mathcal{L}$ is the Lebesgue measurable subsets of $\mathbb{R}^n$, we get one possible integral, but we could just as easily take $\mu=\mu_c$, the counting measure on $\mathbb{R}^n$ (where $\mu(A)$ is equal to the number of points in $A$), and $\mathcal{M}=2^X$, so that $\int_A f d\mu_c= \sum_{x_a \in A}f(x_a)$.

In the context of measure theory, we usually just use the term "integral" to refer to the integral defined above, and talk about the "integral with respect to the Lebesgue measure" when talking about the special case mentioned in your question, though this could also just be called the "integral" when the measure space is clear from context.

Edit: One more thing, the definition you give is the way you get from the integral defined for positive functions to the more general definition, and that isn't really important for your question. The definition you should make sure you understand is the one involving the suprema of integrals of simple functions; the definition you give follows very naturally after you understand this.

  • $\begingroup$ Thanks for clarifying. Does that mean that the Lebesgue Integral could produce two different answers for the same function (if we vary our choice of measure)? $\endgroup$
    – FD_bfa
    Dec 18, 2022 at 22:53
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    $\begingroup$ Yes, but we wouldn't really phrase it like that. The key here is that you have to have the measure space (including the measure itself!) given beforehand, and once this is fixed, the idea of an integral is unambiguous. $\endgroup$ Dec 18, 2022 at 22:58
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    $\begingroup$ The result you're referencing is about the integral with respect to the Lebesgue measure, and I'm willing to bet that in one way or another (perhaps implicitly) your text specified the measure space as $(\mathbb{R}^n,\mathcal{L},m)$. But note that even this is not always true. For a function to be integrable with respect to the Lebesgue integral, it needs to be absolutely integrable, whereas with the improper Riemann integral this is not the case. I'll leave constructing this example to you (hint: think about the alternating harmonic series). $\endgroup$ Dec 18, 2022 at 23:10
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    $\begingroup$ Happy to help. Also, note that my example doesn't give two different values per se, but rather that one exists and the other doesn't. Your statement does indeed hold if we talk about the (original) Riemann integral of a bounded function $f$ on an interval $[a,b]$, but sometimes not when we extend it to the improper integral. $\endgroup$ Dec 18, 2022 at 23:17

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