TITLE: Equality of a supremum of two Expectation
QUESTION [5 upvotes]: I got stuck to prove an equality in a detailed way and I hope, someone could tell to me, how to fix it. Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $\mathcal{K}$ be the set of all probability measure $Q$, which are absolutely continuous with respect to $P$. We are given a r.v. $X$. Suppose there is a $N\in\mathbb{N}$ such that for all $n\ge N$ we have
$$\inf_{Q\in\mathcal{K}}Q(X\ge n)=0$$
I want to verify (in detailed way) the following equality:
$$\sup_{Q\in\mathcal{K}}E_Q[-(X\wedge n)]=\sup_{Q\in\mathcal{K}}E_Q[-(X\wedge N)]\ge -N$$
The inequality is clear, since $-(a\wedge b)=-a\vee -b\ge -b,-a.$ 
We can rewrite the equality as:
$$\inf_{Q\in\mathcal{K}}E_Q[X\wedge n]=\inf_{Q\in\mathcal{K}}E_Q[X\wedge N]$$ 
My idea was: $E_Q[X\wedge n]=E_Q[X\wedge n(\mathbf1_{X\ge n}+\mathbf1_{X< n})]=nQ(X\ge n)+E[(X\wedge n)\mathbf1_{X<n}]$. The inf over the first term is zero, but at the end I did not get the RHS. So how can I prove this in a detailed way? The equality is stated in this paper, see proof of Theorem 5.4 the direction $3)\Rightarrow 1)$.

REPLY [2 votes]: The equality is not true. You just need to consider $\mathcal{K}=\{P,Q\}$ (or its convex hull, if you want to satisfy the assumptions in the paper), where $P$ and $Q$ are such that $X$ $=$ $N-1$ a.s. under $P$, but $E^Q[X]<(N-1)$ with $Q(X>N)>0$.
Then clearly $P(X \geq n)=0$ for any $n \geq N$, while \begin{align*}\inf_{\mathcal{K}} E[X \wedge (N+1)]=E^Q[X \wedge (N+1)]=&E^Q[X \wedge N + 1_{\{N<X\}}(X \wedge (N+1)-N)]\\>&E^Q[X \wedge N]=\inf_{\mathcal{K}} E[X \wedge N].\end{align*}