Let $(X_n)_{n\in\mathbb{N}}$ be a sequence of $\mathbb{R}_+$-valued random variables defined on some probability space $(\Omega,\mathcal{F},\mathbb{P})$. Consider a sequence of sub-sigma algebras $\mathcal{F}_n\subset\mathcal{F}$, for $n\in\mathbb{N}$.
Suppose that $\lim_{n\rightarrow\infty}X_n=x\in\mathbb{R}$ in distribution. Then is it true that $\lim_{n\rightarrow\infty}\mathbb{E}[X_n|\mathcal{F}_n]=x$ in distribution?
In general it need not be true. For example you can cook up random variables $X_n$ such that $X_n \overset{d}{\to}x$ but $\lim_{n\to\infty}\mathbb{E}[X_n] \neq x$. Then taking all $\mathcal{F}_n$ to be trivial $\sigma$-field, one gets $\mathbb{E}[X_n|\mathcal{F}_n] = \mathbb{E}[X_n]$ which does not converge to $x$ in any sense.
But it is true if you assume $X_n$ are bounded by some integrable random variable say $X$. Below the proof is done for this case only.
Let us define $Y_n = X_n-x$. It is equivalent to show that $E[Y_n|\mathcal{F}_n] \overset{P}{\to} 0. $ Note that $Y_n$ are bounded by $Y=X+x$ which is also integrable. By assumption $Y_n \overset{d}{\to} 0$ and hence $Y_n \overset{P}{\to} 0$. First let us show $\mathbb{E}[|Y_n|]\to0$ as $n\to\infty$.
For any $\epsilon > 0\ $, $\ \mathbb{E}[|Y_n|] \leq \epsilon.P(|Y_n|\leq\epsilon) + \int_{|Y_n|>\epsilon}Y \leq \epsilon\ + \int_{|Y_n|>\epsilon}Y $.
So letting $n\to\infty$ and using Lebesgue dominated convergence theorem, we get $\lim\limits_{n\to\infty} \mathbb{E}[|Y_n|] \leq \epsilon$ for any $ \epsilon > 0$. Hence $\mathbb{E}[|Y_n|]\to0$ as $n\to\infty$.
Now for the final part let us fix $\epsilon > 0$. Set $A_n=\{|\mathbb{E}[Y_n|\mathcal{F}_n]|>\epsilon\}$ and $1_{A_n}$ be corresponding indicator function. Note that, $$|\mathbb{E}[Y_n|\mathcal{F}_n]|\,1_{A_n} \geq \epsilon.1_{A_n} $$ and using Jenson inequality $\mathbb{E}[|Y_n||\mathcal{F}_n]\geq|\mathbb{E}[Y_n|\mathcal{F}_n]|$. So we have $$\mathbb{E}[|Y_n||\mathcal{F}_n]\,1_{A_n} \geq \epsilon.1_{A_n} $$ Taking expectaion, $$P(A_n) \leq \frac{1}{\epsilon}\int_{A_n}\mathbb{E}[|Y_n||\mathcal{F}_n] \leq \frac{1}{\epsilon}\mathbb{E}[\mathbb{E}[|Y_n||\mathcal{F}_n]] = \frac{1}{\epsilon}\mathbb{E}[|Y_n|]$$ Finally letting $ n\to\infty$, $\lim_{n\to\infty} P(A_n) = 0$. So $E[Y_n|\mathcal{F}_n] \overset{P}{\to} 0$ and hence $E[Y_n|\mathcal{F}_n] \overset{d}{\to} 0.$