Let $f\in L^1(\mathbb{R})$ where the measure is taken to be the Lebesgue measure. The Fourier transform of $f$ is the function $\hat{f}$ defined as $$\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\xi x}f(x)dx \qquad ,\xi\in \mathbb{R}$$ Show that $\hat{f}$ is well-defined, that is, the integral exists (as a Lebesgue integral)
Can anyone check my attempt? I have mistaken somwhere but I couldn't find it. Thanks for your help.
Attempt: For any $x\in \mathbb{R}$, we have $|e^{-i\xi x}f(x)|=|e^{-i\xi x}||f(x)|=|f(x)|$ ,and since $f\in L^1(\mathbb{R})$, we get $$\int_{-\infty}^{\infty}|e^{-i\xi x}f(x)|dx=\int_{-\infty}^{\infty}|f(x)|dx=C<\infty$$
But \begin{align*}\int_{-\infty}^{\infty}|\hat{f}(\xi)|d\xi & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Big|\int_{-\infty}^{\infty}e^{-i\xi x}f(x)dx\Big| d\xi \\ & \leq \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\Big|e^{-i\xi x}f(x)\Big|dx d\xi \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|C|dx d\xi=\frac{C}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\mathcal{L}(\mathbb{R}) d\xi=\infty \end{align*} where $\mathcal{L}(\mathbb{R})$ is the Lebesgue measure of $\mathbb{R}$.