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I am having trouble with this question. It is not an homework question, I am currently trying to practise different problems for an exam.

Let $f$ be a nonnegative measurable function on $\mathbb R^d$, with

$$m(\{x | f(x)>\lambda\}) = \frac{1}{1+\lambda^2}.$$

For which values of $p$ is $f\in L^p$?

Clearly for $p=1$, you have $\int f = \int_0^{\infty} m(\{x | f(x)>\lambda\}) d\lambda < \infty$.

So $f\in L^1$. But how can I use the sets $\{x | f(x)>\lambda\}$ to compute $\int f^p$ for $p>1$?

Any help would be appreciated! Thank you!

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up vote 2 down vote accepted

Let $F(x,t):=\chi_S(x,t)|$ where $S:=\{(x,t),|g(x)|>t\}$. Using Fubini's theorem for non-negative functions, we have,, if $p\geqslant 1$, $$\int_0^{+\infty}\int_0^{+\infty}F(x,t)dxdt=\int_0^{+\infty}\int_0^{+\infty}F(x,t)dtdx.$$ The LHS is $$\int_0^{+\infty}\lambda\{x, |g(x)|>t\}dt$$ and the RHS is $$\int_0^{+\infty}|g(x)|dx.$$ Using that with $|g(x)|:=|f(x)|^p$ we get what we want.

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I am not sure to see why we get what we want. We don't know anything about $m(\{x | f(x)^p > t\}$. It seems to me that you have only proven the well-known identity $\int g(x) = \int m(\{x | g(x) > t\}$ for any function $g$. – stephen Dec 11 '12 at 20:43
The set $\{f(x)^p>t\}$ is the same as $\{f(x)>t^{1/p}\}$. – Davide Giraudo Dec 11 '12 at 20:43
Oh sorry about my confusion! Thank you! – stephen Dec 11 '12 at 20:45

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