Integration of discontinuous functions

In order to evaluate its Fourier transform, I want to determine whether $$f(x)=\arctan(\frac{1}{x})$$ belongs in $$L^1(\mathbb{R})$$, $$L^2(\mathbb{R})$$ or both. Therefore, we have to check the continuity at $$x=0$$.

One solution I have seen is showing that $$f\not\in L^1(\mathbb{R})$$ because $$\arctan(\frac{1}{x})$$ is discontinuous at that point :

$$$$\lim_{x\to0^-} \arctan(\frac{1}{x}) = \frac{-\pi}{2} \not = \frac{\pi}{2} = \lim_{x\to 0^+} \arctan(\frac{1}{x})$$$$

However, I recall from my classes that

$$$$f\in L^1(\mathbb{R}) \iff \int_{\mathbb{R}} |f(x)| dx \text{ exists and is finite}$$$$

Thus, my question is the following : Shouldn't we check the continuity of $$|f|$$ as opposed to $$f$$, i.e :

$$$$\lim_{x\to0^-} |f(x)| = \lim_{x\to0^+} |f(x)|$$$$

since we're trying to determine the integrability of $$|f|$$ and not $$f$$ ? In this example, the result would differ, as $$|\arctan(\frac{1}{x})|$$ is continuous at $$x=0$$:

$$$$\lim_{x\to0^-} |\arctan(\frac{1}{x})| = |\frac{-\pi}{2}| = \frac{\pi}{2} = |\frac{\pi}{2}| = \lim_{x\to 0^+} |\arctan(\frac{1}{x})|$$$$

• Why do you want to check continuity? To find whether something is in $L^p$ or not, the relevant criterium is, as you stated, that its integral is finite, not that it's continuous. Commented Jan 23 at 19:44

Since $$\{x=0\}$$ is a null set, you don't really need the fact that $$\arctan(\frac{1}{x})$$ is continuous thanks to the properties of the integral.