 Solution

Define $g$ by the inverse integral. Its absolute convergence gives $|g(x)|\le (2\pi {)}^{-d}\Vert \hat{f}{\Vert }_1$. If $x_n\to x$, the exponential factors converge pointwise and their differences are bounded by $2|\hat{f}|$. The dominated convergence theorem proves $g(x_n)\to g(x)$. Therefore $g$ is bounded and continuous and equals $f$ almost everywhere.