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.
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$\begingroup$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.
$\endgroup$ 6 $\begingroup$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 $\sqrt\pi$.
$\endgroup$ 3 $\begingroup$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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