Continuity of a probability integral

Question

Let $m: \mathcal{B}(\mathbb{R}^n) \rightarrow [0,1] $ be a probability measure without point masses.

Let $f:\mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}$ be (jointly) continuous.

Define the mapping $\varphi: \mathbb{R}^m \rightarrow [0,1]$ as $$ \varphi(x) \ := \ m\left( \left\{ v \in \mathbb{R}^n \mid f(v,x)>0\right\} \right) = \int_{\mathbb{R}^n} \mathbb{1}_{ \{ v \mid f(v,x) > 0 \} }(v) \ m(dv) . $$

Questions.

1. Is $\varphi$ continuous?
2. Is $\varphi$ continuous, if we assume $f$ disjointly continuous (i.e. $\forall v $, $f(v,\cdot)$ is continuous and $\forall x$, $f(\cdot,x)$ is continuous)?

Answer 1

I am wondering if the following is a counterexample: $f(v,x) = vx$, with dimensions $n = m = 1$.

Then $\varphi(0) = 0$; while for all $\epsilon>0$, $\varphi(\epsilon) = m( \mathbb{R}_{>0} )$, which may be strictly positive.