Example of function $g(x)$ s.t. $E(g|X_n|) \not\to 0$ Consider the following result (assuming probability space ($\Omega, \mathcal F, \mathsf P$))

If $g : [0, \infty) \to [0,\infty)$ bounded, strictly increasing, $g(x) > 0$ for $x > 0$ and $\lim_{x \to 0^+}g(x) = 0$, then
$$X_n \to 0 \quad \text{in probability} \iff E(g(X_n)) \to 0.$$

I managed to prove this result. I am interested to see what happens if we relax some of the conditions needed for this result. In particular, when we relax boundedness we see that for $X_n(\omega) = n \mathbb {1}_{[0,1/n)}(\omega)$ we have that $X_n \to 0$ in probability but for $g(x) = x$, we see that
$$E(g(X)) = n \mathsf P(\{\omega \in [0,1/n)\}) = n(1/n) = 1 \not\to 0.$$
Here is where my question comes in:

Can we find a $g$ and $(X_n)$ from above satisfying all the conditions except for strictly increasing, such that
$$X_n \to 0 \quad \text{in probability but} \quad E(g(X_n)) \not\to0?$$

 A: Let $g$ be a bounded non-negative function such that $\lim_{x \to 0^+}g(x) = 0$. Then 
$$
\inf_{t\geqslant \varepsilon}g(t)\cdot \Pr\{\left\vert X_n\right\rvert\gt\varepsilon\}\leqslant \mathbb E\left[g\left(\left\vert X_n\right\rvert\right)\right]\leqslant \sup_{t\geqslant 0}g(t)\Pr\{\left\vert X_n\right\rvert\gt\varepsilon\}+\sup_{0\leqslant t\leqslant \varepsilon}g(t)
$$
hence if $g$ is not necessarily increasing  but we instead assume that $\inf_{t\geqslant \varepsilon}g(t)$ is positive for all $\varepsilon\gt 0$, then $\mathbb E\left[g\left(\left\vert X_n\right\rvert\right)\right]\to 0$ is equivalent to the convergence of $(X_n)$ in probability to $0$. 
The positiveness of $\inf_{t\geqslant \varepsilon}g(t)$  for all $\varepsilon\gt 0$ cannot be removed for the implication $\mathbb E\left[g\left(\left\vert X_n\right\rvert\right)\right]\to 0\Rightarrow X_n\to 0$ in probability (but it can for the opposite implication).
Otherwise, let $g$ be defined as $g(x)=x$ on  $[0,1]$, extended by $1$-periodicity on the intervals $(k,k+1)$ and $g(k)$ positive for all integer $k$. Then let $X_n$ be the(constant) random variable equal to $1+1/n$. Since $g(\left\lvert X_n\right\rvert)=1/n$ and $X_n\to 1$ in probability, we get a counter-example.
A: The result is true if g is only bounded and $\lim_{x\to0^+} g(x)=0$, because
$0\le Eg(X_n)=\int_{\{X_n\le\delta\}}g(X_n)+\int_{\{X_n>\delta\}}g(X_n)\le \sup\limits_{0<x<\delta}g(x)+P(X_n>\delta)\sup\limits_{0<t} g$
Using that $\lim_{x\to0^+} g(x)=0$, when $n\to\infty$ we get $Eg(X_n)\to 0$
