Let $m: \mathcal{B}(\mathbb{R}^n) \rightarrow [0,1] $ be a probability measure without point masses and let $f:\mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}$ be continuous. Let $\epsilon > 0$.
Define the mapping $\varphi: \mathbb{R}^m \rightarrow [0,1]$ as $$ \varphi(x) \ := \ m\left( \left\{ v \in \mathbb{R}^n \left| \ \sup_{y \in \{x\} + \epsilon \mathbb{B} } f(v,y)>0\right. \right\} \right). $$
Question: is $\varphi$ continuous?
Not necessarily. Let $n = m = 1$. Let $f(a,b) = b$.
If $x \leq -\epsilon$, $\sup_{y \in \{x\} + \epsilon B} f(\nu,y) \leq 0$ for any $\nu$, hence $\varphi(x) = m(\emptyset) = 0$.
For $x > -\epsilon$, $\sup_{\cdots} f(\nu,y) = b+ \epsilon + x > 0$ for any $\nu$ and hence $\varphi(x) = m(\mathbb{R}) = 1$.