For each $\xi>0$, $S+\xi B$ is a closed bounded set,
where $S$ is a closed bounded subset of $int ~dom f$, $B$ is the unit ball,
$domf$ denotes the effective domain of the proper convex function $f$.
My question is that why the nest of sets
$$ ( S+\xi B ) \cap (int~ dom f)^{C}$$
is empty.
Proof.
If not, there exists $x_0 \in \cap_{\xi>0} H_{\xi}$, where $H_{\xi}= (S+\xi B) ~\cap Q$ with $Q =( int~ domf)^{c} $.
Notice that $S$ is in the interior of domf. For any $y\in S$, there exist $\alpha_{y}>0$ such that $$B(y,0.5\alpha_{y})\subset B(y,\alpha_{y})\subset int ~ dom f.$$ It follows that $$S\subset\cup_{\alpha_{y}>0} B(y,0.5\alpha_{y}).$$ Since $S$ is a closed bounded set of $R^{n}$, $S$ is compact. Thus, $$S\subset \cup_{i=1}^{m} B(y_i,\beta_i),$$ where $\beta_i = 0.5 \alpha_{y_{i}}$ and $m$ is an positive integer. Thus, there exist $\xi_0= \min \{\beta_i\mid i= 1,2,\cdots,m\}$ such that
$$S +\xi_0 B\subset \ int ~dom f .$$
Together with the above inclusion, $x_0 \in H_{\xi_0}$ implies that $$x_0 \in int ~dom f.$$ However, $x_0 \in H_{\xi_0} \subset Q$ which shows us a contradiction.