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