Integral wrt probability measure

Question

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

Answer 1

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}