Is there a function satisfying the following condition? 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. 
 A: 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 
$$A_m=\left\{t: \frac{1}{4^{m+1}}<h(t)\leq \frac{1}{4^{m}} \right\}.$$
Since $h$ is non-decreasing, every non-empty $A_m$ is an interval. We can assume also $h$ is right-continuous in order to avoid the situations when $A_m$ is a set with one point. 
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.
