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If the powers of a function $f$ are Lebesgue integrable what can we say about the original function? for example if we take $f=\frac{1}{x} on [1, \infty] $, it is not integrable but $f^2$ is. Is there is a general condition?

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If $(X, \mu)$ is your measure space, then the set of all $f$ such that $|f|^p$ is integrable (with respect to $\mu$) is denoted $L^p(X,\mu)$. In general, one can always find functions contained in $L^p(X,\mu)$ but not $L^q(X,\mu)$, for any two powers $p,q$. So the answer to your main question is no, in general we cannot relate the integrability of a function to the integrability of its powers.

However, for the case that $(X,\mu)$ is such that $\mu(X) < \infty$, a.k.a. a finite measure space (e.g. a probability space), then one has that $L^q(X) \subseteq L^p(X)$ if $p \le q$. In other words, if a large power of $f$ is integrable, then a smaller power of $f$ will also be integrable.

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You might be interested in the Orlicz spaces. Suppose that $\Phi: \mathbb{R}\rightarrow \mathbb{R}$ is a convex even function. Under various conditions on $\Phi$, $$\{f: \int_\Omega \Phi(f)\, d\mu < \infty\}$$ is a vector space. This vector space can be given a (somewhat unexpected ) norm. Have a look at this link:

http://en.wikipedia.org/wiki/Birnbaum%E2%80%93Orlicz_space

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The first question you should always ask is about where $f$ lives. A sufficient condition is that the measure of the entire space be finite. For general measure spaces, the notions of integrability mean-square integrability are highly distinct - consider both the example you provided and that of $f(x) = x^{- 1 /2} $ on $[0,1].$

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So you are saying for example for $f: \mathbf{X}\rightarrow\mathbf{R}$, it doesn't hold? –  Anita Oct 8 '12 at 1:32
    
I'm saying you have to be more stringent in your conditions on $X$ to guarantee it. –  user17794 Oct 8 '12 at 16:03
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