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

The convergence 
$$\exp\Big(\int_\mathbb{R} (e^{izx}-1)d\lambda_n(x) \Big) \to \exp\Big(\int_\mathbb{R} (e^{izx}-1)d\nu(x)\Big)$$ yields the convergence $$\int_\mathbb{R} (e^{izx}-1)d\lambda_n(x) \to  \int_\mathbb{R} (e^{izx}-1)d\nu(x).$$
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-(Tx)^{-1}\sin(Tx))d\lambda_n(x) \\
&\le& 1+1/\pi,
\end{eqnarray*}
since the function sinc is bounded below by $-1/\pi$. 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+1/\pi.
\end{eqnarray*}
Thus $\nu$ is finite. I guess that a refinement of this argument shows that $\nu(\mathbb{R}) \le 1$.