Does the Fourier inversion theorem hold if $f$ is square-integrable? I know that for $f$ an integrable function, it's Fourier transform may not be integrable, so the inversion theorem does not necessarily hold.
Assume $f$ is square-integrable instead. Does this imply that the inversion theorem holds, i.e. we can recover $f$ by integrating its Fourier transform? If not, what further assumptions are required?
I am getting different information from different notes online. Some say above is enough, others say that we need some continuity properties, and yet others say that the inversion theorem does not hold pointwise but only in $L^2$ sense?
 A: Here's the story. For $f\in L^1(\Bbb R)$ let $\hat f$ be the usual Fourier transform:
$\newcommand\ft{\frac1{\sqrt{2\pi}}}$ 
$$\hat f(\xi)=\ft\int e^{-i\xi t}f(t)\,dt.$$
One can state the Inversion Theorem like so:


Inversion Theorem. If $f,\hat f\in L^1$ then $\hat{\hat f}(x)=f(-x)$.


Now for $f\in L^2$ and $A>0$ define $$F_A(f)=\ft\int_{-A}^Ae^{-i\xi t}f(t)\,dt.$$
The main theorem is


Plancherel  Theorem There exists an isometric isomorphism $F:L^2\to L^2$ such that $\lim_{A\to\infty}\|Ff-F_Af\|_2=0$ for $f\in L^2$; $F$ is almost its own inverse: $FFf(x)=f(-x)$.


Note that if $f\in L^2\cap L^1$ then it's easy to see that $Ff=\hat f$;  this is why $F$ is also called the Fourier transform.
So in particular $F_Af\to Ff$ in norm, not almost everywhere. No. In  fact


Theorem (Carleson) If $f\in L^2$ then $F_Af\to Ff$ almost everywhere.


Proof: Very very hard. A huge big deal in Fourier analysis.


Corollary. If $f\in L^2$ then $Ff\in L^2$ and $F_AFf(x)\to f(-x)$ almost everywhere.


It's not clear to me exactly what it means to say the inversion theorem holds almost everywhere, but it seems pretty likely that the corollary answers your question.
A: The Fourier transform is always defined, but you have to work with tempered distributions. In what follows I will use the following definition of Fourier transform $$\mathcal F(\phi)(\xi) := \int_{\mathbb R^n}\phi(x)\exp(-2\pi i x \xi)dx$$ for $\phi \in L^1(\mathbb R^n)$.
The Fourier transform is also defined for the space $\mathcal S'(\mathbb R^n)$ of tempered distributions, and it is indeed a isomorphism. The Fourier transform in this case is given by 
$$\langle \mathcal F(u),\phi\rangle:= \langle u, \mathcal F(\phi) \rangle,$$
for $\phi \in \mathcal S(\mathbb R^n)$, the Schwartz space of $\mathbb R^n$.
The space $L^2(\mathbb R^n)$ is contained in $\mathcal S'(\mathbb R^n)$, and restricting the Fourier Transform to it we obtain the isomorphism:
$$\mathcal F:  L^2(\mathbb R^n) \to L^2(\mathbb R^n).
$$
Furthermore, this isomorphism is an isometry, that is,
$$\Vert \mathcal F(\phi) \Vert = \Vert \mathcal \phi \Vert.$$
This identity is called Placherel theorem.
For a more detailed discussion about this I recommend you to take a look at Folland's book Real Analysis, and at the Fourier Analysis: An Introduction, by Stein and Shakarchi .
A: Your last sentence is correct: the inversion theorem does not hold pointwise but only almost everywhere, which means that you recover the same function in $L^2$ sense (remember that two functions are considered equal in $L^2$ sense if they only differ an a null set). If you want it to hold everywhere you need some continuity properties. Thus all sources are probably correct; they just differ in what they mean with equality. 
