For a given function $f\in L^1(\mathbb{R})$, suppose that the  

$$\check{f}(x)=\int\_\mathbb{R} \hat{f}(\zeta)e^{2\pi i\zeta x}d\zeta$$
almost every where converges in $\mathbb{R}$. Then, can we say that
$f=\check{f}$ almost everywhere? If the answer is NO, is it possible that the Lebesgue measure of $\{x: f(x)\neq\check{f}(x)\}$ to be infinite?



On the Fourier inversion formula
*Note: I am not sure if I understand the word "converges" correctly.*


This is completely analogous to the similar question regarding convergence of Fourier series, which is classical.


Let $$g(x,r) = \int\_{-r}^r \hat f(\zeta) e^{2\pi i \zeta x} d\zeta$$ by "partial sums" of the inverse Fourier transform, and denote by $$h(x, r) = \int\_0^1 g(x, r t) dt = \int\_{-r}^r \hat f(\zeta) e^{2\pi i \zeta x} (1 - \tfrac{|\zeta|}{r}) d\zeta $$ the Cesàro averages of $g$.


By Plancherel's theorem, $g(\cdot, r)$ is the convolution of $f$ with the function $\phi\_r(x) = 2 r \operatorname{sinc}(\pi r x)$ (which plays the same role as the Dirichlet kernel in the theory of Fourier series). In a similar way, $h(\cdot, r)$ is the convolution of $f$ with an $\psi\_r(x) = r (\operatorname{sinc}(\pi r x))^2$ (which serves as the continuous counterpart of the Fejér kernel).


Since $\psi\_r(x)$ is an approximate identity as $r \to \infty$ (that is: $\psi\_r(x) = r \psi\_1(r x)$, $\psi\_r(x) \ge 0$ and $\int\_{-\infty}^\infty \psi\_r(x) dx = 1$), and additionally $\psi\_1$ is bounded by a "radially decreasing" and integrable function: $\psi\_1(x) \leqslant \min\{1, 1 / (\pi x)^2\}$. This implies that the functions $f \* \psi\_r$ converge to $f$ as $r \to \infty$ almost everywhere (and also in $L^1$); see, for example, Corollary 2.43 in *Advanced Real Analysis* by David McCormick and José Luis Rodrigo, available [here](https://warwick.ac.uk/fac/sci/masdoc/people/masdoc_alumni/davidmccormick/ara-v0.1.pdf). Hence, $h(x, r) \to f(x)$ almost everywhere as $r \to \infty$ (this is stated just below the proof of Corollary 2.43 in the book linked above).


For a fixed $x$, if $g(x, r)$ has a limit as $r \to \infty$, then the limit is necessarily equal to the limit of Cesàro means $h(x, r)$. Thus, if $g(x, r)$ converges for almost all $x$ as $r \to \infty$, then the limit is equal to $f(x)$ almost everywhere.


