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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 $\lim_{\mu(A)\rightarrow 0} (\int_A \gamma) = 0$?

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I would like to hear you elaborate on why this question is "widely applicable to a large audience" :) – ShawnD Mar 13 '12 at 2:15
It's classified... – Adam Mar 13 '12 at 3:58
up vote 2 down vote accepted

In order for this to work, you have to add some additional conditions, if I'm not mistaken. (As always, comments and corrections are welcome.) I shall present some arguments showing that additional conditions are in fact needed. But first, let us examine what kind of additional conditions might work.

Let $g: \mathbb{R}^n \setminus \{0\} \times \mathbb{R}^n \times X \rightarrow \mathbb{R}_{> 0}$ have all the properties specified in the question, and let $g$ also be essentially bounded (bounded, except possibly on a set of measure zero) in the third argument. Then the answer is yes, because an essentially bounded measurable function on a probability space must be integrable, so in this case every function $h$ of your family will have to be measurable and integrable in the third argument, so by the argument presented in Conditions under which the Limit for "Measure $\to 0$" is $0$, $\lim_{\mu(A) \rightarrow 0} \ \int_{A} h(z,y,x ) \mu(dx) = 0$, independently of $\delta$ even.

So, if we add essential boundedness in the third argument, the answer is yes. If for example, $X$ is compact, then boundedness follows from local boundedness, so assuming compactness should be sufficient to answer your question positively. (I do not think it is necessary, however. Neither is essential boundedness.)

We shall now construct a counterexample. (Functions of the form you were thinking about, will indeed suffice to do that.) Let $n=1$ for simplicity and let $X=(0,1)$ equipped with the Lebesgue measure. Define $g: \mathbb{R} \setminus \{0\} \times \mathbb{R} \times (0,1) \rightarrow \mathbb{R}_{> 0}$ as follows:


This function has all the nice properties you seek (continuity, local boundedness and measurability in the appropriate arguments) and is integrable for $z=y$ because $g(z,z,x)=1$ for all triples $(z,z,x)$. If fails to be integrable for $z\neq y$ (independently of $\delta$), because as we know $x\mapsto\frac1x$ is not integrable around $0$. Because of this, for any interval of the form $A=(0,c)$, we must necessarily have $\int_A g(z,y,x)\mu(dx)=\infty$ as long as $z\neq y$, so the supremum in this case will not be $0$ but $\infty$, no matter how close together $y$ and $z$ might get ...

I hope this helps.

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