$E(g(X)), E(g'(X)) <\infty $ implies $\lim_{x\rightarrow \infty} f(x)g(x)= 0$ ($f$ is the density of $X$)?

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I am trying to figure out the Stein's identity which asserts that for r.v $X$ having pdf $$p_\theta(x)=\exp\{ \theta T(x)-A(\theta)\}h(x)$$ where $ T$ is differentiable and $g>0$ is differentiable s.t. $$E(g(X)), E(|g'(X)|) < \infty,$$ then $$E\left(\left[\frac{h'(X)}{h(X)}+\theta T'(X)\right]g(X)\right)=-E(g'(X))$$ provided the support of $X$ is $(-\infty,\infty)$.

The method is integration by part but one needs to show that $$\lim_{x\rightarrow \infty} p_\theta(x)g(x) =0.$$

The condition $E(g(X)) < \infty$ alone seems not enough ( there could be some spikes of $p_\theta g$) but I do not know how to utilize the condition $E(g'(X))< \infty$.

I have also found if $p_\theta g$ is uniform continuous, then $p_\theta g$ would approach 0 but how to show the uniform continuity then?

or $p_\theta g$ may not approach 0?

Any help would be appreciated!