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Let $\mu$ be a probability measure over the (closed but unbounded) set $X \subseteq \mathbb{R}^m$: $\int_X \mu(dx) = 1$.

Consider function $f:\mathbb{R}^n \times \mathbb{R}^n \times X \rightarrow \mathbb{R}_{\geq 0}$ that is:

  • continuous in the first argument;
  • locally bounded in the second argument.

Moreover, function $x \mapsto f(y,y,x) $ is integrable for all $y \in \mathbb{R}^n$.

Consider a sequence $\{z_i\}_{i=1}^{\infty}$ such that: $z_j \in \mathbb{R}^n$, $z_j \rightarrow z$.

I would like to state the following result.

$ \forall \epsilon \in \mathbb{R}_{>0} \quad \exists i^* \in \mathbb{N} $ such that:

$$ \displaystyle \int_{X} | f(z,z_i,x) - f(z_i,z_i,x) | \mu(dx) \ \leq \ \epsilon \qquad \forall i \geq i^*. $$

Proof?

Additional notes: if the result is not true, I'm interested in both finding a counterexample and finding additional assumptions that would lead the result.

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1 Answer 1

up vote 1 down vote accepted

Try $n=1$, $f(x,y,z) = g((y-x)/y)$ for $y \ne 0$, $f(x,0,z) = 0$, where $g$ is a bounded continuous function with $g(1) \ne g(0)$. Then for any sequence $z_i \to 0$ with $z_i \ne 0$, $\int |f(0,z_i, x) - f(z_i, z_i, x)|\ \mu(dx) = |g(1)-g(0)|$.

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Thanks! You might be interested in solving:math.stackexchange.com/questions/116894/… –  Adam Mar 6 '12 at 0:44

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