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.
On
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 $\sqrt\pi$.
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.