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I have a question about measure theory.

Let $(X,\mathcal{F},\mu)$ be a measure space. Let $f$ be a $\mu$-integrable nonnegative function on $X$. I'm looking for a function $\varphi$ satisfying the following:

  • $\varphi: \mathbb{R} \to \mathbb{R}$, nondecreasing,
  • $\lim_{x \to \infty} \varphi(x)/x=\infty$,
  • $\int_{X} \varphi \circ f\,d\mu<\infty$.

My attempt \begin{align*} \int_{X} \varphi \circ f\,d\mu&=\int_{0}^{\infty} \varphi(x)\, \nu(dx)=\int_{0}^{\infty} \nu(\{\varphi>t\})\,dt, \end{align*} where $\nu(A)=\mu(f^{-1}(A))$, $A \in \mathcal{B}(\mathbb{R})$. From the Markov inequality, \begin{align*} \nu(\{\varphi>t\}) \le \frac{1}{t} \int_{0}^{\infty}\varphi \,d \nu. \end{align*} However, $\int_{1}^{\infty} 1/t\,dt=\infty$. Is there sharper upperbound of $\nu(\{\varphi>t\})$ under a suitable condition on $\varphi$? What $\varphi$ should satify?

If you know, please let me know.

1 Answers 1

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One cannon prove convergence of integral for arbitrary $\varphi$. The goal is to construct some $\varphi$ that the integral converges.

We have that $f$ is a $μ$-integrable nonnegative function on $X$, so $\displaystyle\int_X f\,d\mu<\infty$. It implies that $$h(t)=\displaystyle\int_X f\cdot 1_{\{f\geq t\}}\,d\mu \downarrow 0 \text{ as } t\to\infty.$$

Note that $h(0)=\displaystyle\int_X f\,d\mu <\infty$.

Define for $m\in\mathbb Z$ the sets $$A_m=\left\{t: \frac{1}{4^{m+1}}

Let $A_m=[z_m, z_{m+1})$, if this set is non-empty. Note also that $h(0)<\infty$ implies that there exists some $m_0$ such that $0\in A_{m_0}$.

For $t\in A_m$ set $\varphi(t)=2^mt$. This function is such that $\varphi(t)/t\to\infty$ as $t\to\infty$.

Then $$ \int_X \varphi(f)\,d\mu = \sum_{m\geq m_0} \int_X \varphi(f)\cdot 1_{\{f\in A_m\}}\,d\mu = \sum_{m\geq m_0} \int_X 2^mf\cdot 1_{\{f\in A_m\}}\,d\mu = $$ $$ =\sum_{m\geq m_0} 2^m\int_X f\cdot 1_{\{f\in A_m\}}\,d\mu\leq \sum_{m\geq m_0} 2^m\int_X f\cdot 1_{\{f \geq z_m\}}\,d\mu = $$ $$=\sum_{m\geq m_0} 2^m \cdot h(z_m)=\sum_{m\geq m_0} 2^{-m} < \infty. $$

The proof above is copied with some minimal changes from the proof of the same result for expectations of random variables given here: pages 8-9, second part of Lemma 1.