Problem
Assume the probability space $(\Omega,\mathcal{F},P)$ and a non-negative random variable $H$ on $\Omega$ are given. $E_P[H] < \infty$ does not necessarily hold. Now the claim is that there exists a probability measure $Q$ on $\Omega$ that is equivalent to $P$ and under which $E_Q[H] < \infty$ is satisfied. Why is that? (The problem is in connection to showing that an arbitrage-free price for a discounted claim H exists in an arbitrage-free market.)
Approach
I tried to show this by defining a new random variable $Y := c(1+H)^{-1}$ where $c$ is a normalizing constant. $Y$ surely is non-negative and smaller than $c$ hence bounded. Then I defined the measure $Q(A) := \int_A Y dP$ for any $A \in \mathcal{F}$. The Theorem of Radon-Nikodym now implies that $Y = \frac{dQ}{dP}$ and that yields $E_Q[H] = E_P[H\frac{dQ}{dP}] = cE_P[H/(1+H)] \leq c < \infty$. It is clear that $Q \ll P$ but how can I show $P \ll Q$?
Here's a bare-hands approach. Let $B_n:=\{n-1<H\le n\}$ for $n=1,2,\ldots$. If you can make $Q(B_n)\le C\cdot n^{-3}$ then $E_Q[H]$ will be finite. So let's choose the Radon-Nikodym derivative $dQ/dP=c_n$ on $B_n$, with $c_n>0$ to be chosen. We need $\sum_nc_nP(B_n)=1$ for $Q$ to be a probability. A simple choice accommodating both needs is $c_n:=C\cdot n^{-3}/[1+P(B_n)]$, with $C:=\left[\sum_n{P(B_n)\over n^3(1+P(B_n))}\right]^{-1}$.