Consider two sets $L$ and $L^{\prime}$ such that:
$i\notin L$;
$i\notin L^{\prime}$;
some function $\nu:\{L,L^{\prime},L\cup\{i\},L^{\prime}\cup\{i\}\}\longrightarrow \mathbb{R}^{+}$ (where $\mathbb{R}^+$ denotes non-negative reals) such that:
$\nu(L)\leqslant\nu(L^{\prime})$;
$\nu(L\cup\{i\})\leqslant\nu(L^{\prime}\cup\{i\})$;
$\nu(L)\leqslant\nu(L\cup\{i\})$;
$\nu(L^{\prime})\leqslant\nu(L^{\prime}\cup\{i\})$;
$\nu(L)\leqslant\nu(L\cup\{i\})-\eta$, where $\eta\geqslant0$;
Then, is the following statement true?
If Conditions 1 to 7 hold, then $\nu(L\cup\{i\})-\eta\leqslant\nu(L^{\prime})$.
If possible, be so kind to include a brief proof (or a sketch of it). If the statement is true only under some additional (or alternative!) restrictions, please be so kind to state them. If it is false, please disprove it. If the problem is not well-defined or some additional information is missing, let me know!
Thank you all very much for your time!
If $\nu(L)=\nu(L'\cup \{i\})$, then all of the inequalities become equalities and $\eta=0$. So assume $\nu(L)<\nu(L'\cup \{i\})$: then without loss of generality, we can shift and rescale $\nu$ so that $\nu(L)=0$ and $\nu(L'\cup\{i\})=1$. Write $a=\nu(L')$, $b=\nu(L\cup\{i\})$, $x=b-\eta$. Then the information given amounts to
$$\begin{gather} 0\leq a\leq 1 \\ 0\leq b\leq 1 \\ 0\leq x \leq b\text{.} \end{gather}$$
Does this information imply $x\leq a$, or do there exist values $x$, $a$, and $b$ satisfying these hypotheses such that $x>a$? You may want to draw a Hasse diagram.