Return to Answer

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 $\nu(\mathbb{R}) \le 1$.

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

The convergence $$\exp\Big(\int_\mathbb{R} (e^{izx}-1)d\eta_n(x) \Big) \to \exp\Big(\int_\mathbb{R} (e^{izx}-1)d\nu(x)\Big)$$ yields the convergence $$\int_\mathbb{R} (e^{izx}-1)d\eta_n(x) \to \int_\mathbb{R} (e^{izx}-1)d\nu(x).$$ I take the real parts and change the signs to have non-negative functions. $$\int_\mathbb{R} (1-\cos(zx))d\eta_n(x) \to \exp \int_\mathbb{R} (1-\cos(zx))d\nu(x).$$ 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\eta_n(x)\Big)dz \\ &\le& \liminf_n \frac{1}{T}\int_0^T \Big(\int_\mathbb{R} (1-\cos(zx))d\eta_n(x)\Big)dz \\ &=& \liminf_n \int_\mathbb{R} (1-(Tx)^{-1}\sin(Tx))d\eta_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 $\nu(\mathbb{R}) \le 1$.

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*}

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 $\nu(\mathbb{R}) \le 1$.

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 $$\int_\mathbb{R} (e^{izx}-1)d\lambda_n(x) \to \exp \int_\mathbb{R} (e^{izx}-1)d\nu(x).$$ $$\int_\mathbb{R} (1-\cos(zx))d\lambda_n(x) \to \exp \int_\mathbb{R} (1-\cos(zx))d\nu(x).$$ By Fubini's theorem and Fatou's lemma \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*}

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*}