I am quite confused about the notion of integrability. In the context of an introduction to complex analysis and Fourier transforms, I am told that if $f$, a complex or real valued function, satisfies the following:

$$ \int_{-\infty}^{\infty} \lvert f(x) \rvert dx<\infty$$

then it is absolutely integrable. However, does this also imply that $f$ is integrable? What can we conclude about $f$ given the above condition on its absolute value (or modulus). I have no notions of Lebesgue integrability.


3 Answers 3


A function $f$ is Lebesgue integrable iff its integral exists and $|\int f| <\infty$. However, an equivalent definition is the one you have give above.

For $f$ with a defined (Lebesgue) integral by definition we have $\int f = \int f^+ - \int f^-$. If $f$ is in addition integrable it then $\int f^+ <\infty$ and $\int f^- < \infty$ so $\int f^+ + \int f^- = \int |f| < \infty$ so where I have used the fact that $|f| =f^+ + f^ - $. Thus $|f|$ is integrable as all non-negative functions have a defined integral.

Conversely $|f|$ is integrable then $f^+ \leq |f|$ and $f^- \leq |f|$ so $ \int f^+ < \int |f| < \infty $ and $ \int f^- < \int |f| < \infty $ which implies both the negative and positive parts of $f$ are finite and the integral is thus just the difference of two positive real numbers, so is finite. Thus $f$ is integrable.

  • 1
    $\begingroup$ What do you mean by $f^{+}$ and $f^{-}$. Also how does the last line prove that $f$ is integrable? $\endgroup$
    – john melon
    Commented Apr 28, 2017 at 8:03
  • $\begingroup$ $f^+(x) = \begin{cases} f(x), & \text{if $f(x)>0$} \\ 0, & \text{otherwise} \end{cases}$ $f^-(x) = \begin{cases} f(x), & \text{if $f(x)<0$} \\ 0, & \text{otherwise} \end{cases}$ $\endgroup$
    – Bernard W
    Commented Apr 28, 2017 at 8:08
  • $\begingroup$ Ok this makes sense then. Just the last line. I cannot seem to understand the passage where you apply the absolute value again to the integral. $\endgroup$
    – john melon
    Commented Apr 28, 2017 at 8:11
  • 1
    $\begingroup$ I rewrote it to make it a little more intuitive. I have also assumed that you know basic properties of the integral, such as monotonicity and linearity. $\endgroup$
    – Bernard W
    Commented Apr 28, 2017 at 8:15
  • $\begingroup$ thank you! Makes perfect sense now. $\endgroup$
    – john melon
    Commented Apr 28, 2017 at 8:16

Absolutely integrable and Lebesgue integrable are the same. If you want an example of a function which is absolutely integrable but not Riemann integrable, consider $f(x) = \begin{cases} e^{-x^2} & for \ x\in\mathbb{Q} \\ -e^{-x^2} & for \ x\notin\mathbb{Q} \end{cases}$. This function is not Riemann integrable (it's not continuous almost everywhere), but it's absolute value is just $e^{-x^2}$ which has integral $\sqrt\pi$.

  • $\begingroup$ Yes but how is its integral $\sqrt\pi$ in the Rationals? $\endgroup$
    – john melon
    Commented Apr 28, 2017 at 8:02
  • $\begingroup$ It's not. $f$ is a function on real numbers, it's defined for every real number, and it's integral is a real number. It just happens that it is defined in terms of whether the input is rational or not. $\endgroup$
    – Alex Jones
    Commented Apr 28, 2017 at 9:06
  • $\begingroup$ Does 'absolutely integrable' mean 'absolutely integrable in Lebesgue sense'? or in Riemann sense? I am asking this because the Dirichlet function is known to be Lebesgue integrable but not absolutely integrable in Riemann sense. $\endgroup$
    – Mike Park
    Commented Nov 3, 2021 at 8:54

What about the following example?

Let $a<b$ and $f:[a,b]\rightarrow \mathbb{R}$, defined by

\begin{equation*} f(x)=\left\{ \begin{array}{ll} 1, & x\in [a,b]\cap\mathbb{Q}\\ -1, & x\in [a,b]\cap\mathbb{R}\setminus\mathbb{Q}. \end{array} \right. \end{equation*}

$f$ is not Darboux integrable on $[a,b]$, hence, it is not Riemann integrable [a,b]; but $|f|$ is Riemann integrable.


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