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Let $\Theta\subseteq\mathbb{R}^d$ is open set and $(\cal X, \cal A)$ be a measurable space . For every $\theta\in\Theta$, suppose that $P_\theta$ is a probability measure on $(\cal X, \cal A)$. Suppose we have measurable function $J:\cal X\times \Theta\rightarrow\mathbb{R}^d$such that \begin{equation} \int{J(x,\theta)dP_{\theta}(x)}=0,\quad \forall \theta\in\Theta \end{equation} and \begin{equation}\int{|J(X,\theta)|^2dP_{\theta}}<\infty, \quad \forall \theta\in\Theta. \end{equation} where $|.|$ denotes the Euclidean norm. If we know the function $f$ defined by $f(\theta)=\int{J(x,\theta)J(x,\theta)^T}dP_{\theta}$ is continuous at $\theta_0\in\Theta$, then I'd like to prove that \begin{equation} \lim_{a\to\infty}\limsup_{\theta\to\theta_0}\int|J(x,\theta)|^21_{\{|J(x,\theta)|\geq a\}}dP_{\theta}=0. \end{equation} To prove this, I consider the simplest case when $d=1$. We have \begin{align*} L&=\lim_{a\to\infty}\limsup_{\theta\to\theta_0}\int|J(x,\theta)|^21_{\{|J(x,\theta)|\geq a\}}dP_{\theta}\\ &=f(\theta)-\lim_{a\to\infty}\limsup_{\theta\to\theta_0}\int|J(x,\theta)|^21_{\{|J(x,\theta)|<a\}}dP_{\theta}. \end{align*} The first term tends to $f(\theta_0)$ due to the continuity of $f$. I am sure that the last term also tends to $f(\theta_0)$ since $a$ is very big, but I could not justify this mathematically. Could anyone help me please? I am also confused what the relation between $|.|$ and $f$ is when $d>1$. Thank you in advance

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I don't think that you can prove this without adding some continuity conditions on $J$ and $P_\theta$. Your problem smells like uniform integrability. In order to use such things one could try to find a dominating measure $\lambda$ for the family $P_\theta$ and write \begin{equation} f(\theta) =\int|J(x,\theta)|^2 \varphi(\theta,x) d\lambda(x) \end{equation} with the $\lambda$-densities $f(\theta,\cdot)$ of $P_\theta$. –  Jochen Wengenroth Jan 31 '14 at 8:26
However, if you an $\varepsilon$ more integrability like $\int |J(x,\theta)|^{2+\varepsilon}dP_{\theta}(x) \le C$ for $\theta$ close to $\theta_0$ then the almost trivial estimate $\int Y^2 I_{\{|Y|\ge a\}}dP \le a^{-\epsilon} \int |Y|^{2+\epsilon}dP$ helps. –  Jochen Wengenroth Jan 31 '14 at 8:34

1 Answer 1

up vote 2 down vote accepted

Here is (I believe) a counterexample:

For $\theta\in\Theta=(-1,1)$ let $$P_\theta= \frac{\theta^2}{2} \delta_{-1/\theta} +\frac{\theta^2}{2} \delta_{1/\theta}+(1-\theta^2)\delta_0$$ (where $\delta_x$ is the Dirac measure in $x$) if $\theta\neq 0$ and $P_0=(\delta_{-1}+\delta_1)/2$. For $J(\theta,x)= x$ your function $f$ is constant and for $\theta_0=0$ you have \begin{equation} \lim_{a\to\infty}\limsup_{\theta\to\theta_0}\int|J(x,\theta)|^21_{\{|J(x,\theta)|\geq a\}}dP_{\theta}(x)=1. \end{equation}

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Thank you very much for your help. I'll modify my assumption in this problem. –  Jlamprong Jan 31 '14 at 11:20
If you think that this answers your question you could accept the answer and then ask another one. –  Jochen Wengenroth Jan 31 '14 at 12:34

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