 Uniqueness of the Fourier transform on integrable functions

For integrable functions on the real line, equality of their Fourier transforms implies equality almost everywhere. To prove it, subtract and assume $\hat{h}=0$. Convolve $h$ with a Gaussian heat kernel $q_t$. The convolution theorem gives $\widehat{h\ast q_t}=0$, which is integrable, so the Fourier inversion theorem gives $h\ast q_t=0$ almost everywhere. Finally $q_t$ is an approximate identity: $\Vert h\ast q_t-h{\Vert }_1\le \int q_t(y)\Vert h(\cdot -y)-h{\Vert }_{1}dy\to 0$. This follows from translation continuity in Lp with exponent one and the vanishing Gaussian tails. Hence $h=0$ in $L^1$. If the two original functions are continuous, equality almost everywhere strengthens to equality everywhere, since a nonzero continuous difference has magnitude bounded below by a positive constant on some interval.