Condition in a theorem in Probability theory. I passed by a simple theorem in Probability theory , yet it really bugs me that I think that 1 condition in the hypothesis is not necessary. 
After checking the proof for many times, I still can't understand author's intention .
Can any one help me by giving me a proper answer?
*The Theorem.
"Suppose $X_n \rightarrow X$ a.s Let g,h be continuous functions with:
i) $g \ge 0 ; g(x) \rightarrow  \infty $ as $|x| \rightarrow \infty$ 
ii) $\frac{|h(x)|}{g(x)} \rightarrow 0$ as$ |x| \rightarrow \infty$, and,
iii) $Eg(X_n)\le K < \infty \forall n$
Then $Eh(X_n) \rightarrow Eh(X)$"
*My Question
Is  the condition  "$g(x) \rightarrow  \infty $ as $|x| \rightarrow \infty$" necessary? Why?
*The provided Proof.

-Durett R.,Probabability Theory and Examples, 2010, page 30
Thank you in advance

The author uses one property which is implied by $g(x) \to \infty$ for $|x| \to \infty$, namely that there is $M>0$ with $g(x) > 0$ for all $|x| \geq M$.
But apart from that, this assumption is not used.
But observe that if $g$ is bounded (this is not the same as $g(x) \not \to \infty$ for $|x|\to\infty$, though), the claim is more or less trivial, because with (ii) this yields that $h$ is bounded, so that
$$
\Bbb{E}(h(X_n)) \to \Bbb{E}(h(X))
$$
follows trivially by continuity of $h$ and dominated convergence.
