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



Integral wrt probability measure
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}


