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I am looking at the following proof from Casella and Berger's Statistical Inference. However, I don't understand the final statement. The only condition on $g'$ here is that $E|g'(X)|<\infty$. But how does this ensure that $g(x)e^{-(x-\theta)^2/(2\sigma^2)} \to 0$ as $x \to \pm \infty$?

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1 Answer

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The hypotheses are that, for some nonnegative functions $f$ and $h$, the function $fh$ is integrable on $(0,+\infty)$, that the function $h$ is decreasing on some $(\theta,+\infty)$ and that $h(x)\to0$ when $x\to+\infty$, and the question is whether all this implies that $g(x)h(x)\to0$ when $x\to+\infty$, where $$g(x)=c+\int_0^xf(y)dy$$ To prove this is true, fix some $z>\theta$ and note that, for every $x>z$, $$g(x)h(x)=g(z)h(x)+h(x)\int_z^xf(y)dy\leqslant g(z)h(x)+\int_z^xh(y)f(y)dy$$ Since $h(x)\to0$ when $x\to+\infty$, this yields $$\limsup_{x\to\infty}g(x)h(x)\leqslant\int_z^\infty h(y)f(y)dy$$ When $z\to\infty$, the RHS can be made as small as desired hence $$\limsup_{x\to\infty}g(x)h(x)=0$$ which proves the result.

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