Set-Up
Let $f$ be a tempered distribution given by integration against a function (which we also call $f$). Suppose that there is a function $g \in L^2$ such that $\mathcal{F}f = g$. In other words, the tempered distribution $\mathcal{F}f$ is given by integration against an $L^2$ function.
Question
Do we have $$ \int |f|^2 = \int |g|^2 ? $$
Observations
$\mathcal{F}^{-1}(g)$ is an $L^2$ function, so it suffices to prove $f(x)=\mathcal{F}^{-1}(g)(x)$ as functions for a.e. $x$.
$\int f \psi = \int \mathcal{F}^{-1}(g) \psi$ for all Schwartz functions $\psi$, so that $f = \mathcal{F}^{-1}(g)$ as tempered distributions.
If $f$ and $g$ are continuous bounded functions, then $$ f(x) = \lim_{\epsilon \to 0} f \ast \psi_{\epsilon} (x) = \lim_{\epsilon \to 0} g \ast \psi_{\epsilon} (x) = g(x) \quad \text{for all $x$}, $$ where $\psi_{\epsilon}$ is a Schwartz approximation to the identity.
Sub-Question If two functions are equal as tempered distributions, under what conditions are they equal (or equal a.e.) as functions?
Special Case
This special case is connected to Convolution theorem with distributions
Take $f = u \ast \varphi$, where $\varphi \in S$ (Schwartz space) and $u \in S'$ (space of tempered distributions).
With appropriate conditions on $u$, the tempered distribution $\mathcal{F}(u \ast \varphi)$ is integration against a function $g \in L^2$.
Note $u \ast \varphi$ is the function defined by $$ (u \ast \varphi)(x) = \langle u, R \tau_x \varphi \rangle, $$ where $\tau_x \phi(y)=\phi(y-x)$ and $R\tau_x\phi(y) = \phi(x-y)$. In the particular case where $u$ is a function, $$ (u \ast \varphi)(x) = \int u(y) \phi(x-y) dx. $$ Note also $u \ast \varphi$ is a $C^{\infty}$ function and a tempered distribution.