# Can a function be square integrable without being integrable?

Reading Tolstov's 'Fourier Series', which states that $f(x)$ is square integrable if both $f$ and its square both have finite integrals over some interval. I haven't seen this restriction on $f$ before, which makes me wonder - can squaring a function ever turn a diverging integral into a converging one?

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As long as the measure space is of infinite measure, this ca happen: Consider $\frac{1}{x}$ on $(1,\infty)$.

If the measure space is finite, this cant happen by Cauchy Schwarz (with the constant 1 function as the second factor).

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Thanks, should've specificed 'finite interval'. –  Benjamin Lindqvist Jun 7 '14 at 18:34
So requiring $f(x)$ to be integrable is actually redundant when talking about square integrability over a finite interval? –  Benjamin Lindqvist Jun 7 '14 at 18:36
The usual definition of square integrablity does not include the requirement that the function itself be integrable. –  Vladimir Jun 7 '14 at 18:39
But note that you still have to require $f$ to be measurable (if we are talking about Lebesgue integrals), consider e.g. $f = \chi_V - \chi_{V^c}$ on a finite measure space with $V$ not measurable. Then $f^2 \equiv 1$ is measurable although $f$ is not. You can do the same (even simpler) with $f = \chi_{\Bbb{Q} \cap I} - \chi_{I \setminus \Bbb{Q}}$ in the case of the Riemann integral. –  PhoemueX Jun 7 '14 at 19:09

The same way the square of $\frac{1}{n}$ converges, but itself does not converge.

So for functions $\frac{1}{x}$ is an example over $[1,\infty)$.

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