Counterexample $X_n \to X$ in distribution but $\lim_{n \to \infty} E[ \log(1+X_n) ]\neq E[ \log(1+X) ]$ Let $X_n \to X$ in distribution where we only consider non-negative random variables.   
I am looking for a counterexample that for the followig limit
\begin{align}
\lim_{n \to \infty} E[ \log(1+X_n) ]=  E[ \log(1+X) ].
\end{align}
Here is a counterexample that I have created. Let $X_n$ have a probability mass function according to $P[ X_n=0]=(1-\frac{1}{n})$ and  $P[ X_n= 2^n]=\frac{1}{n}$. 
Then
\begin{align}
\lim_{n \to \infty} E[ \log(1+X_n) ]= \lim_{n \to \infty}  \frac{1}{n} \log(1+2^n) = \log(2).
\end{align}
On the other hand, we have that  $X_n \to X= 0$ in distribution so 
\begin{align}
E[ \log(1+X) ]=  \log(1)=0.
\end{align}
My question:  Can we create a counterexample such that  $\sup_{n} E[X_n]<\infty$ and $E[X]<\infty$? Also, is there such an example? 
Note that in my counterexample, we have that
\begin{align}
\sup_{n} E[X_n]= \sup_{n} \frac{1}{n} 2^n=\infty.
\end{align} 
 A: No, there is no such counterexample. If we assume additionally that $\sup_n \mathbb{E}(X_n)<\infty$ and $\mathbb{E}(X)< \infty$, then
$$\lim_{n \to \infty} \mathbb{E}\log(1+X_n) = \mathbb{E}\log(1+X).$$
Proof: Choose $\chi_k \in C_b$ such that $0 \leq \chi_k \leq 1$, $\chi_k|_{B(0,k)}=1$ and $\chi_k|_{B(0,k+1)}=0$. Clearly,
$$|\mathbb{E}(\log(1+X_n))-\mathbb{E}(\log(1+X))| \leq I_1+I_2+I_3$$
where
$$\begin{align*} I_1 &:= \sup_{n \geq 1} |\mathbb{E}(\log(1+X_n)-\log(1+X_n) \chi_k(X_n))| \\ I_2 &:= |\mathbb{E}(\log(1+X_n) \chi_k(X_n)-\log(1+X) \chi_k(X))| \\ I_3 &:= |\mathbb{E}(\log(1+X)-\log(1+X) \chi_k(X))| \end{align*}$$
Convergence in distribution implies that
$$\lim_{n \to \infty} \mathbb{E}(\log(1+X_n) \chi_k(X_n)) = \mathbb{E}(\log(1+X) \chi_k(X))$$
for any $k \in \mathbb{N}$, and so $I_2 \to 0$ as $n \to \infty$. On the other hand, we have
$$\begin{align*} |\mathbb{E}(\log(1+X_n) \chi_k(X_n)) - \mathbb{E}\log(1+X_n)| &\leq \mathbb{E}(\log(1+X_n) 1_{|X_n|>k})  \\ &\leq c_k \mathbb{E}(X_n) \end{align*}$$
for some $c_k>0$ such that $c_k \to 0$ as $k \to \infty$. (Use that $x \mapsto x$ is growing much faster than $x \mapsto \log(1+x)$ for large $x$.) The same estimate holds with $X_n$ replaced by $X$. Combining the estimates we get
$$\limsup_{n \to \infty} |\mathbb{E}(\log(1+X_n))-\mathbb{E}(\log(1+X))| \leq c_k \sup_{m \in \mathbb{N}} \mathbb{E}(X_m) + c_k \mathbb{E}(X).$$
Since the right-hand side converges to $0$ as $k \to \infty$ we conclude that
$$\lim_{n \to \infty} \mathbb{E}\log(1+X_n) = \mathbb{E}\log(1+X).$$
