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A doubt: With Lebesgue measure, $\int_{\mathbb{R}\setminus \left\{0\right\} } \frac{1}{x^2}dx$ converges?

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    $\begingroup$ Can't this be decided just using improper Riemann integral on $(0,1)$? [We have an antiderivative $-1/x$ which when evaluated from $a$ to $1$ and then make $a \to 0^+$ diverges.] $\endgroup$ – coffeemath Jan 12 at 4:01