Weak $L_1$ convergence Given a sequence $Y_{un}$, where $Y_{1n},Y_{2n},\ldots$ have the same domain. Assume for every $u\in \mathbb{N}$ we have $e^{itY_{un}}\rightarrow \mathbb{E}[e^{it M}]$ weakly in $L_1$ as $n\rightarrow \infty$. For the defintion see below.
Zhoraster gave an useful example under which i was needed to update the requirements.
Assume for the sequence of random variables $X_n$ it holds
$$
\lim_{u\rightarrow \infty} \limsup_{n\rightarrow \infty}\mathbb{P}\left(d\left(X_n,Y_{un}\right)\geq \epsilon\right)=0 \tag1
$$
Can we conclude that $e^{itX_n}\rightarrow \mathbb{E}\left[e^{itM}\right]$ weakly in $L_1$ as $n\rightarrow \infty$?
I think this should be possible:
Due to (1)  we have for every $n\in \mathbb{N}$ that $\mathbb{E}\left|e^{itX_{n}}-e^{itY_{un}}\right|\rightarrow 0$ as $u\rightarrow \infty$.
And one could formally use $Y_{un}\xrightarrow{n \rightarrow \infty }M_{u}:=M\xrightarrow{u\rightarrow \infty} M$.
We could then see this result as an improved version of Billingsley 1968 Theorem 4.2
If this is not the case: what about the condition (1) replaced by
$$
\lim_{u\rightarrow \infty} P\left(\sup_{n\in \mathbb{N}} d\left(X_n,Y_{un}\right)\geq \epsilon\right)=0 \tag{2}
$$

Definition: Given a sequence of rv $Z_n$ on $(\Omega,\mathcal{F},P)$ such that for all bounded $\mathcal{F}$-mb random variables $K$ it holds
  $$
\lim_{n\rightarrow \infty} \mathbb{E}\left[Z_nK\right]=\mathbb{E}\left[ZK\right],
$$
  we say $Z_n\rightarrow Z$ weakly in $L_1$.

 A: As it was hinted in the comments, the result is not true unless you assume uniform convergence somewhere. 
Here is a simple example. Let $X_n = X$ have non-degenerate Bernoulli distribution and $Y_{un} = X \mathbf{1}_{u\ge n}$. Then $P(Y_{un}\neq X_n)\to 0$, $u\to\infty$, and $e^{itY_{un}} \to 1$, $n\to\infty$, in all senses. However, $e^{itX_n} = e^{itX}\not\to 1$, $n\to\infty$, in any sense.

With the extra assumption (1), this is simple. 
Write for any bounded $K$ and for arbitrary $u\ge 1$, $\varepsilon>0$ by the triangle inequality
$$
\left|\mathbb{E}[e^{itX_n}K] - \mathbb{E}[e^{itM}K]\right|\le \left|\mathbb{E}[e^{itX_n}K] - \mathbb{E}[e^{itY_{un}}K]\right| + \left|\mathbb{E}[e^{itY_{un}}K] - \mathbb{E}[e^{itM}K]\right|\\
\le \sup|K|\cdot\Big(\mathbb{E}\left[\left|e^{itX_n}-e^{itY_{un}}\right|\mathbf{1}_{|X_n-Y_{un}|>\varepsilon}\right] + \mathbb{E}\left[\left|e^{itX_n}-e^{itY_{un}}\right|\mathbf{1}_{|X_n-Y_{un}|\le\varepsilon}\right]\Big) \\
+ \left|\mathbb{E}[e^{itY_{un}}K] - \mathbb{E}[e^{itM}K]\right| \le \Big\{ |e^{ix}-e^{iy}|\le |x-y|\Big\}\\
\le \sup|K|\cdot \Big(2\mathbb{P}\left(|X_n-Y_{un}|>\varepsilon\right) + \varepsilon\Big) + \left|\mathbb{E}[e^{itY_{un}}K] - \mathbb{E}[e^{itM}K]\right|.
$$
Letting $n\to\infty$, we get
$$
\limsup_{n\to\infty}\left|\mathbb{E}[e^{itX_n}K] - \mathbb{E}[e^{itM}K]\right|\le \sup|K|\cdot  \Big(2\limsup_{n\to\infty} \mathbb{P}\left(|X_n-Y_{un}|>\varepsilon\right) + \varepsilon\Big).
$$
The claim now follows by first letting $u\to\infty$ and then $\varepsilon\to 0$.
