For a fixed $f \in L^1(\Omega)$, find a coercive function $h$ such that $h(f) \in L^1(\Omega)$. I am looking for a proof of the following statement :
Let $(\Omega,\mathcal{A},\mu)$ a probability space and $f \in L^1(\Omega)$.
Show that there exists a coercive function $h$ ($h$ is positive, non-decreasing, with $\underset{t \rightarrow +\infty}{\lim} \frac{h(t)}{t} = + \infty$) such that $h(f) \in L^1(\Omega)$.
If i am looking at a simple space like the space of real sequences equipped with the counting measure, it means that if a sequence of positive numbers $a_n$ verifies $$\sum_n a_n< +\infty$$
I can find $h$ coercive such that
$$\sum_n h(a_n)< +\infty$$
which i guess look quite possible by choosing $h$ such that $h(a_n)=\lambda_n a_n$ with $\lambda _n>0$ and $\lambda_n \underset{n \rightarrow +\infty}{\longrightarrow} + \infty$, but again I'm unsure how to detail that.
Any advices are welcomed.
 A: Edit: So it's not because I don't appreciate upvotes, but the solution below is wrong, I realized. Indeed, really it "proves" that any function in $L^p$ will also be in $L^{p+\delta}$ for suitably small ($f$-dependent) $\delta$. This is wrong, though.
Indeed, the mistake made is that I ended up applying monotone convergence to a decreasing net of functions and that does not fly. Take for instance the sequence $f_n(x)=\frac{1}{nx}$ on $[0,1]$. None of them are in $L^1$, but their limit sure is.
However, I did manage to find a proof.
Note that we may, without loss of generality, assume that $f$ is strictly positive. For general $f$ you can simply define $h$ that works for $f^+$ and set $\tilde{h}(x)=0$ for $x\leq 0$.
Thus,
$$
\int_{\Omega} h(f)\textrm{d}\mu=\int_{\Omega} \frac{h(f)}{f} f\textrm{d}\mu=\int_{(0,\infty)} \frac{h(x)}{x}\textrm{d}\nu, 
$$
where $\nu$ is the finite measure on $\mathbb{R}$ given by the push-forward of the measure $f\textrm{d}\mu$ under $f$. Thus, writing $g(x)=\frac{h(x)}{x}$, we've rephrased the problem as: Given a finite measure $\nu$ on $(0,\infty)$, find a positive $g\in L^1(\nu)$ such that $\lim_{x\to\infty} g(x)=\infty$ and such that $g(x)x$ is non-decreasing. We can, without loss of generality, assume that $\nu$ is a probability measure and that $\nu$ has unbounded support (the bounded support case is trivial).
Define $b_0=0$ and recursively, define $b_{k+1}=\sup\{x\in (b_k,\infty)| \nu((b_k,x))\leq 2^{-(k+1)}\}$. Next, define $g(x)=\sum_{k=1}^{\infty} k1_{(b_{k-1},b_k)}$ and note that
$$
\int_{(0,\infty)} g(x)\textrm{d}x= \sum_{k=1}^{\infty} k \mu((b_{k-1},b_k))\leq \sum_{k=1}^{\infty} k 2^{-k}<\infty
$$
Furthermore, since $g$ is non-decreasing, $xg(x)$ will also be non-decreasing. Since $\nu$ has unbounded support, we have that $g$ is increasing and all in all, we get that $h(x)=xg(x)$ is a coercive function such that $\int_{\Omega} h(f)\textrm{d}\mu<\infty$.
Wrong solution:
Well, thinking of sequences does not exploit the strongest assumption that you've been given: That you're working on a probability space.
Define $h_{\alpha}(t)=(\max\{t,1\})^{\alpha},$ which is clearly coercive for any $\alpha>1$.
Note that, $\alpha\mapsto h_{\alpha}(t)$ is non-decreasing for every $t$. Applying monotone convergence, we get
$$
\int_{\Omega} \max\{f,1\} \textrm{d}\mu=\int_{\Omega} \lim_{\alpha\to 1} h_{\alpha}(f)\textrm{d}\mu=\lim_{\alpha\to 1} \int_{\Omega} h_{\alpha}(f)\textrm{d}\mu,
$$
Now, $\int_{\Omega} \max\{f,1\}\textrm{d}\mu\leq \int_{\Omega} |f|\textrm{d}\mu+\mu(\Omega)<\infty,$ so we must have that $\int_{\Omega} h_{\alpha}(f)\textrm{d}\mu<\infty$ for $\alpha$ sufficiently small.
This gives you the desired.
