The formula is:
$f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-i\lambda x}\hat{f}(\lambda)d\lambda$
where $\hat{f}$ is the characteristic function, $f$ is continuous bounded on $R$ and both $f,\hat{f}$ are integrable.
Proof:
$\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-i\lambda x}\hat{f}(\lambda)d\lambda = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-i\lambda x}(\int_{-\infty}^{\infty}e^{i\lambda y}f(y)dy) d\lambda $
At this point, the proof says we cannot change the order of integration and hence we introduce a regularizing factor $g(\epsilon \lambda)=e^{-\lambda^2\epsilon^2/2}$ by:
$\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-i\lambda x}\hat{f}(\lambda)d\lambda = lim_{\epsilon\rightarrow 0}\int_{-\infty}^{\infty}e^{-i\lambda x}\hat{f}(\lambda)g(\epsilon \lambda)d\lambda$
And then he was able to change the order of integration. So I do not understand why we had to introduce $g(\epsilon \lambda)$ because if $f$ is integrable, $e^{i\lambda(y-x)}f(y)$ would be integrable over $R^2$ and can change the order of integration, right? Appreciate a hint, thanks!