Let $\mu$ be a probability measure on $X$, so that $\int_X \mu(dx) = 1$.
In Conditions under which the Limit for "Measure $\to 0$" is $0$ it is shown that $f: X \rightarrow \mathbb{R}_{\geq 0}$ measurable and integrable guarantees $ \lim_{\mu(A) \rightarrow 0 } \int_A f(x) \mu(dx) = 0 $
Now let $g: \mathbb{R}^n \setminus \{0\} \times \mathbb{R}^n \times X \rightarrow \mathbb{R}_{> 0}$ be continuous in the first argument, locally bounded in the second, locally bounded and measurable in the third.
Moreover $\forall z \in \mathbb{R}^n \setminus \{0\}$ $ \quad x \mapsto g(z,z,x)$ is integrable.
I would like to say that $\exists \delta > 0$ such that for the family
$ \mathcal{F} \doteq \left\{ g(z,y,x) \mid \ y \in \mathbb{B}(z,\delta) \right\} $
we have the following fact:
$\forall z \in \mathbb{R}^n \setminus \{0\} \qquad \sup_{h \in \mathcal{F}} \ \lim_{\mu(A) \rightarrow 0} \ \int_{A} h(z,y,x ) \mu(dx) = 0 $
Is it true?
Notes: $\mathbb{B}(z,\delta)$ is the closed ball centered at $z$ and with radius $\delta$.
I'm thinking on functions like $\gamma(z,y,x) = \lambda(z,x) + |z-y| \cdot \Lambda(x)$, where $\lambda(z,x)$ is integrable $\forall z$, while $\Lambda(x)$ is not. Therefore $\gamma$ is integrable if $y=z$ and it is not integrable even if $y = z \pm \epsilon$. But can examples like this destroy the property of having $\lim_{\mu(A)\rightarrow 0} (\int_A \gamma) = 0$?