This doubt is born because I am reading an article in this [link in pag 12](https://arxiv.org/pdf/1505.05116.pdf) in order to use these ideas to prove another problem that raised me. My doubt is following:


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.** 



Let $\mathcal{M}(\Xi)$ set of all probability destributions on $\Xi$. Supremum over $\mathcal{M}(\Xi)$ is equal to sup over Dirac distributions
The attempt you made to prove inequality $ \leq $ is correct. The other desiguladad is consequence of the following general fact:



> 
> For every fixed function $g$ and every probability distribution $\mu$,
>  $$\int\_Xgd\mu\leqslant\sup\_X g\cdot\mu(X)=\sup\_Xg$$
> 
> 
> 


