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Suppose we have a function $h$ which is continuous and Lebesgue integrable on $\mathbb{R}$.

We have $f(x)=\frac{1}{2 \pi} \int_{\mathbb{R}} e^{-iux} h(u) du $, for all real $x$.

If $f$ is Lebesgue integrable, then show that $h(u)=\int_{\mathbb{R}} e^{iux} f(x) dx$ , for real $u$.

It is mentioned in the text that it is a standard Fourier transform result, but I want to prove it using Fubini-Tonelli theorem.

Can anyone help?

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    $\begingroup$ This is the Fourier Inversion Theorem. It cannot be proved with a simple application of Fubini's Theorem. $\endgroup$ – Kavi Rama Murthy Jun 3 at 7:18
  • $\begingroup$ The title doesn't really make sense. $\endgroup$ – zhw. Jun 3 at 15:08

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