# Does absolute integrability imply integrability?

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.

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.

• What do you mean by $f^{+}$ and $f^{-}$. Also how does the last line prove that $f$ is integrable? – john melon Apr 28 '17 at 8:03
• $f^+(x) = \begin{cases} f(x), & \text{if$f(x)>0$} \\ 0, & \text{otherwise} \end{cases}$ $f^-(x) = \begin{cases} f(x), & \text{if$f(x)<0$} \\ 0, & \text{otherwise} \end{cases}$ – Bernard W Apr 28 '17 at 8:08
• 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. – john melon Apr 28 '17 at 8:11
• 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. – Bernard W Apr 28 '17 at 8:15
• thank you! Makes perfect sense now. – john melon Apr 28 '17 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 $\sqrt\pi$.

• Yes but how is its integral $\sqrt\pi$ in the Rationals? – john melon Apr 28 '17 at 8:02
• 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. – AlexanderJ93 Apr 28 '17 at 9:06

Let $$a and $$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.