$X_{1}, X_{2}, \ldots .$ are nonnegative $\mathrm{RVs}$, such that $ \boldsymbol{X}_{n} \stackrel{\text { a.s. }}{\rightarrow} \mathbf{0}$ (a.s stands for almost sure convergence) and $\lim\limits _{n \rightarrow \infty} \mathbb{E}\left\{\boldsymbol{X}_{\boldsymbol{n}}\right\}=\mathbf{2}$, where $\mathbb{E}$ stands for expectation of a random variable RV. I want to obtain $$ \lim _{n \rightarrow \infty} \mathbb{E}\left\{\left|1-X_{n}\right|\right\}, \text { if it exists } $$ There is no independence supposed here, I tried to go from the a.s convergence definition or simply by performing upper-bounds but I still fail and can't seem to see the use of the expectation convergence to $2$ which is obviously important here.
Any suggestions?
Note that for $M\geq 1$, $$ \mathbb{E}[|1-X_n|]=\mathbb{E}[(X_n -1)1_{X_n\geq M}]+\mathbb{E}[|1-X_n| 1_{|X_n|\leq M}] $$ Now, for uniformly bounded variables, a.s. convergence implies $L^1$-convergence by the dominated convergence theorem. Thus,
$$ \lim_{n\to\infty} \mathbb{E}[|1-X_n| 1_{|X_n|\leq M}]=1 $$ and $$ \mathbb{E}[(X_n-1)1_{X_n\geq M}]=\mathbb{E}[X_n 1_{X_n\geq M}]-\mathbb{P}[X_n\geq M] $$ since $X_n\to 0$ a.s., we also have $X_n\to 0$ in probability and hence, $$ \lim_{n\to\infty}\left(\mathbb{E}[X_n 1_{X_n \geq M}]-\mathbb{P}[X_n\geq M]\right)=\lim_{n\to\infty}\mathbb{E}[X_n 1_{X_n\geq M}]=\lim_{n\to\infty}\mathbb{E}[X_n]-\lim_{n\to\infty}\mathbb{E}[X_n1_{X_n<M}]=2. $$
Hence, $$ \lim_{n\to\infty}\mathbb{E}[|1-X_n|]=3. $$