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}\intJ(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}\intJ(x,\theta)^21_{\{J(x,\theta)\geq a\}}dP_{\theta}\\ &=f(\theta)\lim_{a\to\infty}\limsup_{\theta\to\theta_0}\intJ(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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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}\intJ(x,\theta)^21_{\{J(x,\theta)\geq a\}}dP_{\theta}(x)=1. \end{equation} 

