Let $(\Xi,\mathcal{E})$ be a measurable space and $\mathbb{Q}_{i}$ measures in this space and $\xi_{i}\in\Xi$ for $i=1,2,\ldots,N$. Let $f:\Xi\rightarrow \mathbb{R}$ be a measurable function.
The question: I need show that $$\sup_{\mathbb{Q}_{i}\in\mathbb{M}(\Xi)}\left[\sum_{i=1}^{N}\int_{\Xi}f(\xi-\xi_{i})\mathbb{Q}_{i}(d\xi)\right]=\sum_{i=1}^{N}\sup_{\xi\in\Xi}f(\xi-\xi_{i})$$ where $\mathcal{M}(\Xi)$ is the set of all probability distributions supported on $\Xi$
My attempt: The first thing I thought is this \begin{align} \sup_{\mathbb{Q}_{i}\in\mathbb{M}(\Xi)}\left[\sum_{i=1}^{N}\int_{\Xi}f(\xi-\xi_{i})\mathbb{Q}_{i}(d\xi)\right] &= \sup\left\{\left.\sum_{i=1}^{N}\int_{\Xi}f(\xi-\xi_{i})\mathbb{Q}_{i}(d\xi) \right| \mathbb{Q}_{i}\in\mathcal{M}(\Xi) \mbox{ for }i=1,\ldots,N \right\} \\ &= \sum_{i=1}^{N}\sup\left\{\left.\int_{\Xi}f(\xi-\xi_{i})\mathbb{Q}_{i}(d\xi) \right| \mathbb{Q}_{i}\in\mathcal{M}(\Xi) \right\}\\ (1)\qquad &\geq \sum_{i=1}^{N}\sup\left\{\left.\int_{\Xi}f(\zeta-\xi_{i})\delta_{\xi}(d\zeta) \right| \xi\in\Xi \right\} \\ &= \sum_{i=1}^{N}\sup_{\xi\in\Xi}f(\xi-\xi_{i}) \end{align}
where $(1)$ follows from the fact that $\mathcal{M}(\Xi)$ contains all the Dirac distributions supported on $\Xi$.
The probblem is the inequality $\leq$, I need a suggestion to prove this inequality.