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

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?

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