$f$ and $\hat f$ are integrable, right-hand limit and left-hand limit exist, then two limits equal.

Question

I've been struggling with some problems about "Fourier Series" and "Fourier Integral".

My main problem is : " The function $f \in L^1(\mathbb R)$ and its fourier integral $\hat f \in L^1(\mathbb R)$ are given, the left-hand limit and the right-hand limit exist on $x=0$, show that two limits equal. "

My trial was using "Inverse Fourier Transfrom". Since $\hat f$ is continuous function and $L^1$-integrable, $f(x)= \frac{1}{2\pi}\int_{\mathbb R} \hat f(\alpha) e^{i\alpha x} d\alpha$ a.e. holds.

The problem says there exist left-hand limit and right-hand limit on x=0, then $e^{i\alpha x} \rightarrow 1 $ as x goes to 0, so the left-hand limit and right-hand one would be the same.

But, my solution have a huge problem. The "inverse" equality only holds "a.e." Only if $f$ is continous, the equality holds everywhere. However, there is no condition about it. Then, how I could solve this problem? Is there any idea solving this problem, without using "the Inverse Fourier Transform"?

Answer 1

If $E$ has measure $0$ then $E^{c}$ is dense in $\mathbb R$. Take $E$ to be the set of points where $f(x)= \frac{1}{2\pi}\int_{\mathbb R} \hat f(\alpha) e^{i\alpha x} d\alpha$ fails. The right hand and left hand limits are equal to limits through sequences from the dense set $E^{c}$. Hence, (by DCT) $f(0+)=f(0-)$.