I hope I did not make a mistake, but I think it works. 

The convergences 
$$\exp\Big(\int_\mathbb{R} (e^{izx}-1)d\lambda_n(x) \Big) \to \exp\Big(\int_\mathbb{R} (e^{izx}-1)d\nu(x)\Big)$$ yields the convergence $$\int_\mathbb{R} (e^{izx}-1)d\lambda_n(x) \to \exp \int_\mathbb{R} (e^{izx}-1)d\nu(x).$$
I take the real parts and change the signs to have non-negative functions. 
$$\int_\mathbb{R} (1-\cos(zx))d\lambda_n(x) \to \exp \int_\mathbb{R} (1-\cos(zx))d\nu(x).$$
By Fubini's theorem and Fatou's lemma, for every $T>0$,
\begin{eqnarray*}
\int_\mathbb{R} (1-(Tx)^{-1}\sin(Tx))d\nu(x) 
&=& \frac{1}{T}\int_0^T\Big(\int_\mathbb{R} (1-\cos(zx))d\nu(x)\Big)dz \\
&=& \frac{1}{T} \int_0^T \lim_n\Big(\int_\mathbb{R} (1-\cos(zx))d\lambda_n(x)\Big)dz \\
&\le& \liminf_n \frac{1}{T}\int_0^T \Big(\int_\mathbb{R} (1-\cos(zx))d\lambda_n(x)\Big)dz \\
&=& \liminf_n \int_\mathbb{R} (1-T^{-1}\sin(Tx))d\lambda_n(x) \\
&\le& \liminf_n \int_\mathbb{R} (1+T^{-1})d\lambda_n(x) = 1+T^{-1}
\end{eqnarray*}
Applying Fatou's lemma again, 
\begin{eqnarray*}
\int_\mathbb{R} 1d\nu(x) 
&\le& \liminf_{T \to +\infty} \int_\mathbb{R} (1-(Tx)^{-1}\sin(Tx))d\nu(x) \\
&\le& \liminf_{T \to +\infty} 1+T^{-1} = 1.
\end{eqnarray*}