Let $(\Omega, \mathcal F,\mu)$ be a measurable space. Let $f:\Omega\rightarrow \mathbb R$ be $\mathcal F$-measurable.

We know that: $\int_\Omega |f|d\mu<\infty\implies\int_\Omega fd\mu<\infty $.

If $\int_\Omega fd\mu<\infty$, then this means the function is integrable, and both $\int_\Omega f^+d\mu<\infty$ and $\int_\Omega f^-d\mu<\infty$, which implies that $|f|=f^++f^-$ is also integrable.

If the reverse holds, then both conditions are equivalent, then why do we use the absolute value. Why not spare notation?

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    $\begingroup$ Often the Lebesgue integral of $f$ is not even defined unless $f\geq 0$ or $|f|$ is integrable. $\endgroup$ Apr 29, 2016 at 17:06

1 Answer 1


Yes the reverse holds as well. This is really a definitional convention. We typically define "$f$ is integrable" to mean that both $\int_\Omega f^+ d\mu$ and $\int_\Omega f^- d\mu$ are finite (since, in the beginning, we define integration only for non-negative functions) then define $$\int_\Omega f d\mu = \int_\Omega f^+ d\mu - \int_\Omega f^- d\mu.$$ However, if both these are finite, one can easily see that $$\int_\Omega \lvert f \rvert d\mu = \int_\Omega f^+ d\mu + \int_\Omega f^- d\mu$$ and so $\int_\Omega \lvert f \rvert d\mu$ is finite as well.

Edit: @Vladamir has a much better answer in the comments. THe order in which we define things matters here. The value $\int_\Omega \lvert f \rvert d\mu$ is always defined (for measureable $f$). The value $\int_\Omega f \, d\mu$ is only defined when the other value is finite. Thus to check $f$ is integrable, we need to check $\int_\Omega \lvert f \rvert d\mu < \infty$ because a priori, the value $\int_\Omega f \, d\mu$ is not defined.

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    $\begingroup$ I always see that a condition for integrability is $\int_\Omega |f|d\mu<\infty$. But this is exactly the same as $\int_\Omega fd\mu<\infty$. Then why the extra "$|\cdot|$"? $\endgroup$
    – telemaco
    Apr 29, 2016 at 17:10
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    $\begingroup$ @julian.marr What if $f$ is complex valued? That's where absolute value is truly needed. $\endgroup$
    – layman
    Apr 29, 2016 at 17:13
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    $\begingroup$ One cannot write $\int f d\mu<\infty$ if $\int f d\mu$ is not defined in the first place. On the other hand, if $f$ is $\mu$-measurable, then the integral $\int|f|d\mu$ is always defined (but can be infinite), and if it is finite, then $f$ is integrable and the integral of $f$ is well defined. $\endgroup$
    – Vladimir
    Apr 29, 2016 at 17:16
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    $\begingroup$ @julian.marr Sorry for my late reply. If $f$ is a complex valued function, then we can write $f(x) = u(x) + i v(x)$ for real valued functions $u$ and $v$ ($u$ is called the real part of $f$, and $v$ is called the imaginary part). We define $\int f \,d\mu = \int u \,d\mu + i \int v \,d\mu$. When should this be integrable? The right hand side is a complex number. We don't have an ordering $<$ on the complex numbers to say that a complex number $a + bi < \infty$. So when should we say $\int f \,d\mu$ is integrable? Well, we can take the absolute value of a complex number. $\endgroup$
    – layman
    Apr 29, 2016 at 18:38
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    $\begingroup$ @julian.marr $|a + bi|$ is defined to be $\sqrt{a^{2} + b^{2}}$, which is a real number. So $|f|$ is a real valued function if $f$ is complex valued. So we can talk about $\int |f| \,d\mu$, and this is a real number now, so we can talk about when it's finite. $\endgroup$
    – layman
    Apr 29, 2016 at 18:39

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