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

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"?

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) $$f(0+)=f(0-)$$.