1
$\begingroup$

I imagine that the following question has a well known (and perhaps, easily obtainable) answer, but I can't find it by myself nor along the references that I have in mind so far.

So, if $f$ is a nonnegative measurable function defined on some measure space $(\Omega,\mathcal{F},\mu)$ and $\lambda_{f}:[0,+\infty)\to\mathbb [0,+\infty]$ is the distribution function of $f$, namely $$\lambda_{f}(t)=\mu(f^{-1}(t,+\infty))$$ then for all $p>0$ $$\int f^{p} d\mu=p\int_{(0,+\infty)} t^{p-1}\lambda_{f}(t)dt$$ where $dt$ is Lebesgue measure and both integrals are infinite if $f\notin L^{p}_{\mu}$ (see for instance Folland's Real Analysis: Modern Techniques and their Applications, section 6.4).

This is the question: is the statement $f\in L^{p}_{\mu}$ equivalent to $$\sum_{n\geq 1} n^{p-1}\lambda_f(n)<\infty?$$ Remarks: 1. If you want, assume that $f$ is finitely supported. This is: $\lambda_{f}(0)<\infty$.

  1. Since it is not clear to me whether $t\mapsto t^{p-1}\lambda_{f}(t)$ is decreasing (unless $0<p\leq 1$) I do not know whether the integral test applies.
$\endgroup$

1 Answer 1

0
$\begingroup$

Note that if $\mu(X)=\infty$, then your series condition doesn't necessarily imply $f\in L_{\mu}^{p}(X)$. For example, take $X=\mathbb{R}$ with Borel $\sigma$-algebra and Lebesgue measure. If $f=1/2\chi_{[0,\infty)}$, then clearly $f\notin L^{p}$; however, $\lambda_{f}(n)=0$ for $n\geq 1$.


Assume that $\lambda_{f}(0)<\infty$. Observe that for $p>0$,

\begin{align*} \int_{0}^{\infty}t^{p-1}\lambda_{f}(t)\mathrm{d}t=\sum_{n=0}^{\infty}\int_{n}^{n+1}t^{p-1}\lambda_{f}(t)\mathrm{d}t, \tag{1} \end{align*} where we allow the possibility that both sides are infinite.

If $p\geq 1$, then it follows from (1) that

\begin{align*} \sum_{n=0}^{\infty}n^{p-1}\lambda_{f}(n+1)\leq \int_{0}^{\infty}t^{p-1}\lambda_{f}(t)\mathrm{d}t\leq\sum_{n=0}^{\infty}(n+1)^{p-1}\lambda_{f}(n) \tag{2} \end{align*} Noting that $n\sim n+1$ as $n\rightarrow\infty$, completes the proof in this case.

If $0<p<1$, then \begin{align*} \sum_{n=0}^{\infty}(n+1)^{p-1}\lambda_{f}(n+1)&\leq\int_{0}^{\infty}t^{p-1}\lambda_{f}(t)\mathrm{d}t \tag{3}\\ &\leq\lambda_{f}(0)\int_{0}^{1}t^{p-1}\mathrm{d}t+\sum_{n=1}^{\infty}n^{p-1}\lambda_{f}(n)\\ &=p^{-1}\lambda_{f}(0)+\sum_{n=1}^{\infty}n^{p-1}\lambda_{f}(n)\\ \end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.