Does this condition about two measures $p$ and $q$ imply existence of $p$-integrable function that is not $q$-integrable?

Question

Context :

I am trying to characterize some properties of barycenters of measures on a probability space, the following question arises. More precisely I want to restrict the support of a measure on some simplex averaging to some probability measure $p$ to those measure $q$ that satisfy property 1 bellow, it would help to have other characterizations for it to understand it more deeply.

Setup :

Let $p$ and $q$ be two probability distributions over some measurable space $(X,\mathcal X)$, we use the extended real line $\overline{\mathbb R}=[-\infty,\infty]$. I want to relate the three following statements, I think they might be equivalent

1. There exists $\lambda>0$ such that for all $A\in\mathcal X$, $(1+\lambda)p(A)-\lambda q(A)\geq 0$.
2. All $p$-integrable function $f:X\to \overline{\mathbb R}$ is also $q$-integrable.
3. $q\ll p$ and the Radon-Nikodym derivative $\frac{\partial q}{\partial p}$ is $p$-surely bounded by some constant $C$.

I could show that 1 $\Leftrightarrow$ 3 and that 1,3 $\Rightarrow$ 2 (see below), so I am wondering if 2 is actually equivalent to the rest. Example 1 below provide with one illustration of when this is true but not completely trivial.

Examples :

Example 1 : If $X=\mathbb N$ and $\mathcal X=\mathcal P(\mathbb N)$ with $p(n)=\frac{2}{3} 3^{-n}$ and $q(n)=\frac{1}{2}\cdot 2^{-n}$, then the condition is satisfied, indeed it is equivalently rewritten as $\frac{1}{\frac{q(A)}{p(A)}-1}<\lambda$, but clearly the left term tends toward $0$ if we let $A=\{ n\}$. In that case we can use $f(n)=a^n$ for any $2\leq a<3$ to get that $f$ is $p$-integrable and not $q$-integrable.

Attempt:

• 3 $\Rightarrow$ 2 : Suppose that $f$ is $p$-integrable, we get $\int f~dq= \int f\cdot\frac{\partial q}{\partial p}~dq\leq C \cdot\int f~dp<\infty$.
• 1 $\Rightarrow$ 3 : It is clear that $q\ll p$. Rewriting 1, gives existence of $\lambda > 0$ such that for all $A\in\mathcal X$ such that $p(A)>0$, we have $\frac{q(A)}{p(A)}\leq \frac{1+\lambda}{\lambda}$ which implies that $\frac{\partial q}{\partial p}\leq\frac{1+\lambda}{\lambda}$.
• 3 $\Rightarrow$ 1 : Assume WLOG that $C>1$, pick $\lambda=\frac{1}{C-1}$ to get $\lambda q(A) \leq \frac{C}{C-1} p(A)=\left(\frac{1 + C-1}{C-1} \right) p(A)=(\lambda+1) p(A)$
• 2 $\Rightarrow$ 1 : Suppose that 1 is false, equivalently that there exists a sequence $\{ A_n \}_{n\in \mathbb N}$ such that $p(A_n)>0$ and $\frac{q(A_n)}{p(A_n)}\to\infty$, I cannot conclude from this.

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Any help in proving this, any source about such subjects or any comment is most welcome.

Answer 1

I think the following may provide a proof for 2 $\Rightarrow$ 1.

Assume that 1 is false, therefore for all $k\geq 1$ there is $A_k\in\mathcal X$ such that $k\leq \frac{q(A_k)}{p(A_k)}$. Let $f$ be defined as \begin{align*} f(x) &= \sum_{k\geq 1} \frac{\mathbf 1_{A_k}}{p(A_k) k^2} \end{align*} Then by Fatou's lemma \begin{align*} \int f~dp&\leq \liminf_{n\to\infty} \int \sum_{k=1}^n \frac{\mathbf 1_{A_k}}{p(A_k) k^2} ~dp\\ &=\liminf_{n\to\infty} \sum_{k=1}^n \frac{1}{k^2}\\ &<\infty \end{align*} but for any $n\geq 1$ \begin{align*} \int f~dq &\geq \int \sum_{k=1}^n\frac{\mathbf 1_{A_k}}{p(A_k) k^2} ~dq\\ &= \sum_{k=1}^n \frac{1}{k^2}\cdot\frac{q(A_k)}{p(A_k)}\\ &\geq \sum_{k=1}^n\frac{1}{k} \end{align*} Therefore $\int f~dq\geq \sup_{n\geq 1} \sum_{k=1}^n\frac{1}{k}=\sum_{k\geq 1} \frac1k=\infty$.

Therefore $f$ is $p$-integrable and not $q$-integrable, i.e. 2 is false.

Any proof checking or improvement is welcome.